Q1: Describe the problems faced by the Weimar Republic. Ans: After Germany lost World War I, Emperor Wilhelm II abdicated, and the Weimar Republic was formed. The Republic faced the following problems:
Versailles Treaty: The treaty that ended the war, took German territories, and resources, and required 6 billion pounds in reparations, making the Republic unpopular.
Economic Crisis: War debts depleted gold reserves, causing the German mark to fall and prices to soar.
Political Flaws: The Weimar Republic had a weak constitution with issues like proportional representation and Article 48, which allowed the President to rule by decree during emergencies. This made it hard for the Republic to handle economic crises effectively.
Q2: Discuss why Nazism became popular in Germany by 1930. Ans: The end of World War I brought significant changes to Germany’s political landscape, and the newly formed Weimar Republic faced many challenges right from the start.
One of the biggest problems was the Great Depression. The Weimar Republic did little to address the economic crisis, and this created an opportunity for Hitler to present himself as a saviour to the German people, who were suffering from economic hardship and political instability.
Hitler’s powerful speeches promised to build a strong nation, reverse the injustices of the Versailles Treaty, restore the dignity of the German people, and create jobs for everyone. These promises gave people hope.
Nazi propaganda was also very effective. The use of red banners with the Swastika, the Nazi salute, and enthusiastic applause during rallies made Nazism increasingly popular among the people.
Q3: What are the peculiar features of Nazi thinking? Ans: The Nazi thinking was synonymous with Hitler’s worldview. The features of such thinking are as follows:
Nazi ideology ranked races with Nordic Germans at the top and Jews at the bottom, while others were placed in between.
They wanted more land (Lebensraum) for Germany, even if it meant taking it from others.
Nazis twisted Darwin’s theory to justify that only the strongest races should survive.
They valued ‘racial purity,’ deeming those with disabilities and certain ethnic groups as undesirable and subject to extermination.
They saw war as a means to prove German superiority and rejected peace as weak.
Q4: Explain why Nazi propaganda was effective in creating a hatred for Jews. Ans: The Nazi propaganda was highly effective in creating hatred for the Jews due to the following reasons:
The Nazis used language and media very carefully to spread their ideas. They claimed Jews were inferior and undesirable.
They took advantage of old Christian hatred of Jews, who were blamed for killing Christ, to increase German prejudice against them.
Nazis taught children to hate Jews from a young age, firing Jewish teachers and expelling Jewish students.
They made propaganda films, like “The Eternal Jew,” to reinforce negative stereotypes and hatred of Jews.
Such kinds of propaganda made the Germans believe that the Jews were inferior to them.
Q5: Explain the role of women in Nazi society. Return to Chapter 1 on the French Revolution. Write a paragraph comparing and contrasting the role of women in the two periods. Ans:
In Nazi society, women were expected to be housewives, have many pure-blooded children, and avoid ‘undesirables.’
Those who complied received benefits like hospital perks and discounts. However, women who didn’t follow these rules faced severe punishment, including public shame, loss of family, jail, or death.
This was very different from women’s roles during the French Revolution. French women fought for education, equal wages, and freedom from forced marriage. They could pursue careers, become artists, run businesses, and own property.
Q6: In what ways did the Nazi state seek to establish total control over its people? Ans: The Nazis established control over its people by various means:
The Nazis used posters and films to promote their ideas and gain support.
They carefully controlled the media to shape public opinion and stir up hatred against those they labelled ‘undesirable.’
They created special security forces to enforce their rules and used police power to act without limits.
Genocide and fear helped them maintain tight control over society.
Q1. What were the social, economic, and political conditions in Russia before 1905? Ans: The following were the social, economic, and political conditions in Russia before 1905: (a) Social Conditions
The majority religion was Russian Orthodox Christianity— which had grown out of the Greek Orthodox Church. But the empire also included Catholics, Protestants, Muslims, and Buddhists.
Cross of the Russian Orthodox Church
The non-Russian nationalities were not treated equally to the Russian nationalities.
Workers were a divided group on the basis of skill and training.
(b) Economic Conditions
The majority of Russians were agriculturists.
Industries were few. Much of the production was done by the craftsmen.
With the expansion of the Russian rail network, foreign investment in factories grew.
There was huge coal, iron, and steel production. The workers were exploited by capitalists who made their lives miserable.
(c) Political Conditions
Russia was a monarchy.
The Tsars believed in the divine rights of kings.
They were not responsible to the Parliament.
All political parties were illegal in Russia.
Q2. In what ways was the working population in Russia different from other countries in Europe, before 1917? Ans:
Compared to people in Europe, the situation for Russians, especially workers like farmers and factory workers,was very poor.
This was mainly because of Tsar Nicholas II’s harsh and corrupt rule.
Much of what the peasants produced went to landowners and the wealthy.
Large estates were controlled by the nobility, the crown, and the Orthodox Church.
In Europe, peasants supported and fought for nobles, but in Russia, they wanted the nobles’ land for themselves.
They often refused to pay rent and sometimes even killed landlords due to frustration with harsh policies.
Factory workers also suffered; they couldn’t form unions or political groups to voice their complaints.
They were exploited by industrialists, often receiving low wages and working long hours with no limits.
Q3. Why did the Tsarist autocracy collapse in 1917? Ans:
There was widespread anger against the Tsar’s autocratic rule. The Russian people wanted an end to the war and peace, but the Tsar, who had mishandled the war, insisted on continuing it. The peasants, in particular, wanted land for themselves, but the corrupt government ignored their demands.
Workers also faced terrible conditions. They couldn’t form trade unions or political parties to voice their concerns. Private industrialists ran most industries, and workers often received less than minimum wages and worked 12-15 hours a day with no set limits.
The Tsar’s rule had become ineffective. He was corrupt, oppressive, and indifferent to the people’s welfare.
Karl Marx’s ideas inspired people to demand change.
The Tsar’s involvement and defeat in World War I was the final blow that led to the collapse of his rule.
Q4. Make two lists: one with the main events and the effects of the February Revolution and the other with the main events and effects of the October Revolution. Write a paragraph on who was involved in each, who the leaders were, and what the impact of each was on Soviet history. Ans: (a) The February Revolution
In February 1917, there were severe food shortages in the workers’ areas. Parliamentarians were against the Tsar’s decision to dissolve the Duma.
On 22nd February, a factory lockout happened, with many women leading strikes. The strikes spread, with workers surrounding important areas in Petrograd, like Nevskii Prospekt.
On 25th February, the Duma was dissolved, leading to the ransacking of the Police Headquarters on 27th February. The cavalry refused to shoot at the protesters. By evening, soldiers and workers joined together to form the Petrograd Soviet, a “council.”
The Tsar abdicated his power on 2nd March, and the Soviet and Duma leaders set up a Provisional Government for Russia. The February Revolution was led by the people, not any political party. Petrograd was crucial in ending the monarchy and had an important place in Soviet history.
February Revolution
(b) The October Revolution
The conflict between the Provisional Government and the Bolsheviks led to the October Revolution.
In September 1917, Vladimir Lenin began organizing an uprising. On 16th October, he convinced the Petrograd Soviet and the Bolshevik Party to support a socialist takeover. Leon Trotsky led the Military Revolutionary Committee to carry out the plan.
On 24th October, as the uprising began, Prime Minister Kerensky left the city for reinforcements. The Military Revolutionary Committee quickly took action: the ship Aurora shelled the Winter Palace, and by evening, the city was under Bolshevik control.
After fierce fighting, the Bolsheviks secured the Moscow-Petrograd area. Their actions were approved by the All-Russian Congress of Soviets. The October Revolution, led by Lenin and Trotsky, marked the start of Lenin’s rule with Bolshevik leadership.
Q5. What were the main changes brought about by the Bolsheviks immediately after the October Revolution? Ans:
Industries and banks were nationalized, meaning the government took over their ownership and management.
Land was declared social property, allowing peasants to take land from the nobility. In cities, the Bolsheviks divided large houses to fit family needs.
The use of old aristocratic titles was banned, and new uniforms for the army and officials were introduced. The Bolshevik Party was renamed the Russian Communist Party (Bolshevik).
Despite opposition from their allies, the Bolsheviks made peace with Germany and withdrew from World War I.
Later, the Bolsheviks were the only party in elections to the All-Russian Congress of Soviets, which became Russia’s Parliament.
Q6. Write a few lines to show what you know about (a) Kulaks (b) The Duma (c) Women workers Between 1900 and 1930 (d) The Liberals (e) Stalin’s Collectivisation Programme Ans: (a) Kulaks
They were the well-to-do peasants who were supposed to be holding stocks in the hope of higher prices.
They were raided so that they may be eliminated in order to develop modern farms and establish state-controlled large farms.
(b) The Duma The Duma was a consultative parliament that was created with the permission of the Tsar during the 1905 Revolution. (c) Women Workers Between 1900 and 1930
Women made up 31% of the factory labour force but were paid between 1/2 and 3/4 of a man’s wages.
They actively led the strikes in many factories. They even worked in the collective farms.
(d) The Liberals
They were a group that looked to change society. They wanted a nation that tolerated all religions and opposed the uncontrolled power of dynastic rulers.
They argued for a representative, elected parliamentary government with laws interpreted by a well-trained, independent judiciary that was separate from rulers and officials.
(e) Stalin’s Collectivisation Programme
Stalin hoped to solve the problem of food shortage by combining small farms with large and modern farms.
This was a collectivisation programme that began in 1929. Peasants were forced to work in these state-controlled collective farms called Kolkhoz.
Q1. Describe the circumstances leading to the outbreak of revolutionary protest in France? Ans: Several factors contributed to the revolutionary protests in France in 1789:
Social Inequality: French society was divided into three estates. The clergy (First Estate) and the nobility (Second Estate) enjoyed privileges, while the commoners (Third Estate) bore heavy taxes and had no say in political matters. This inequality fueled anger among the Third Estate. American War of Independence
Economic Crisis: France faced a severe financial crisis due to costly wars (like the American War of Independence) and extravagant spending by the monarchy. This led to increased taxation and widespread poverty.
Enlightenment Ideas: Enlightenment philosophers like Rousseau criticized absolute monarchy and advocated ideas of liberty, equality, and fraternity. Their writings inspired revolutionary thinking.
Weak Monarchy: King Louis XVI was seen as an ineffective leader who failed to address the grievances of the people. His inability to reform taxation and government deepened the crisis.
Famine and Unemployment: Bad harvests in the 1780s led to food shortages and inflation, worsening the economic conditions. Widespread unemployment in urban areas further fueled dissatisfaction.
Q2. Which groups of French society benefited from the revolution? Which groups were forced to relinquish power? Which sections of society would have been disappointed with the outcome of the revolution? Ans:
The French society was divided into three estates before the revolution:
The Three Estates
Beneficiaries: The Third Estate, which included peasants, artisans, small farmers, landless labourers, servants, businessmen, merchants, court officials, and lawyers, benefited the most from the revolution. They gained rights such as freedom from feudal dues and the abolition of privileges enjoyed by the nobility and clergy.
Groups that lost power: The First Estate (clergy) and Second Estate (nobility) were forced to relinquish power. These estates had long enjoyed privileges by birth, including exemption from taxes and special political influence. After the revolution, these privileges were removed.
Disappointed groups: Women and the poorer working class were disappointed with the revolution’s outcomes. Women were not granted equal rights or the right to vote, and the poor continued to face economic hardships, as the revolution did not fully address their needs.
Q3. Describe the legacy of the French Revolution for the peoples of the world during the nineteenth and twentieth centuries. Ans:
The French Revolution left a lasting impact on the world during the 19th and 20th centuries:
Liberty, Equality & Fraternity
Spread of revolutionary ideas: The ideals of liberty, equality, and fraternity inspired people across Europe and other parts of the world. The revolution led to the abolition of feudal systems in many countries during the 19th century.
Influence on independence movements: Colonized people adapted the ideas of freedom and equality to fight for independence. The concept of a sovereign nation-state became a powerful idea that motivated many to resist colonial rule.
Influence on notable figures: Leaders such as Tipu Sultan and Raja Ram Mohan Roy were inspired by the revolutionary ideas of France, which encouraged reforms in their own societies.
Q4. Draw up a list of democratic rights we enjoy today whose origins could be traced to the French Revolution. Ans: Many democratic rights we enjoy today can be traced back to the French Revolution:
Freedom of speech
Freedom of expression
Freedom of the press
Abolition of censorship
Right to vote
Abolition of slavery
Right to liberty
Right to property
Right to security
Right to education
Divorce laws
These rights form the foundation of modern democracies, where equality and justice are fundamental principles.
Q5. Would you agree with the view that the message of universal rights was beset with contradictions? Explain. Ans: Yes, the message of universal rights during the French Revolution was indeed marked by contradictions:
Exclusion of women: While the revolution promoted equality, women were excluded from many basic rights. They were not allowed to vote, hold public office, or have a say in the law-making process.
Limited reach of rights: The rights to liberty, property, and security were not extended to women, and they were denied equal opportunities based on their abilities. This created a significant gap between the ideals of the revolution and its implementation.
Thus, while the revolution pushed for universal rights, it failed to include everyone equally, particularly women and certain marginalized groups.
Q6. How would you explain the rise of Napoleon? Ans: The rise of Napoleon Bonaparte can be attributed to several factors:
Napoleon Bonaparte
Political instability: After the revolution, France was left in a state of political turmoil. The weak and unstable government under the Directory created an opportunity for a strong leader to emerge.
Military success: Napoleon’s military victories made him a national hero. His successful campaigns in Italy and Egypt boosted his popularity and increased his influence in French politics.
Support from the army: Napoleon had the loyalty of the French army, which helped him seize power through a coup in 1799, overthrowing the Directory and establishing himself as the ruler.
Introduction of reforms: As emperor, Napoleon introduced reforms such as the protection of private property and the establishment of a uniform system of weights and measures. His legal reforms, known as the Napoleonic Code, carried forward revolutionary principles of equality and modern governance.
Though Napoleon’s reign ended with his defeat at the Battle of Waterloo, his impact on France and Europe remained, with many of his reforms continuing to influence modern societies.
Q1: What do we get from cereals, pulses, fruits, and vegetables?
Ans: The things we get from cereals, pulses, fruits, and vegetables are as follows:
Cereals are the source of carbohydrates and are the main source of energy.
Pulses provide protein for growth and development.
Vegetables and fruits are loaded with minerals, vitamins, carbohydrates, proteins and fats for overall development.
Page No. 142
Q1: How do biotic and abiotic factors affect crop production?
Ans: Two major factors that affect the crop are:
Biotic factors like insects, rodents, pests, and many more spread the disease and reduce crop production.
Abiotic factors like humidity, temperature, moisture, wind, rain, flood and many more destroy the crop raised.
Q2: What are the desirable agronomic characteristics for crop improvement?
Ans: The essential agronomic features required for crop improvement are:
Profuse branching along with tallness in any fodder crop.
Dwarfness in any cereals.
Page No. 143
Q1: What are the macro-nutrients, and why are they called macro-nutrients?
Ans: Macro-nutrients are the fundamental elements that are used by plants in larger quantities. Macro-nutrients needed by the plants are
Macro-nutrients as the constituents of protoplasm.
Phosphorus, Nitrogen, and sulfur are present in proteins.
Calcium exists in the cell wall.
Magnesium is a significant component of chlorophyll.
Q2: How do plants get nutrients?
Ans: Plants get carbon from the air, oxygen from the air and water, and nutrients like nitrogen from the soil.
Carbon: Plants take in carbon dioxide (CO₂) from the air through tiny holes in their leaves and use sunlight to make food (photosynthesis).
Oxygen: They absorb oxygen from the air for respiration and also get it from water through their roots.
Nutrients: Roots absorb essential nutrients like nitrogen, phosphorus, and potassium from the soil to help the plant grow strong and healthy.
Page No. 144
Q1: Compare the use of manure and fertilisers in maintaining soil fertility.
Ans:
Manure improves soil quality with added nutrients.
Manure provides extra organic matter called humus to the soil, therefore increasing the water retention capacity of sandy soils and drainage in clayey soil.
Manures reduce soil erosion.
They provide food for soil-friendly bacteria, which are helpful in growing crops.
The effects of fertilisers are
Fertilisers make the soil too dry and powdery and raise the rate of soil erosion.
The organic matter decreases by decreasing the porosity of the soil; hence, the plant roots do not get enough oxygen.
The nature of soil changes, either basic or acidic.
Page No. 145
Q1: Which of the following conditions will give the most benefits? Why? (a) Farmers use high-quality seeds, do not adopt irrigation or use fertilisers. (b) Farmers use ordinary seeds, adopt irrigation, use fertilisers and use crop protection measures. (c) Farmers use quality seeds, adopt irrigation, use fertiliser and use crop protection measures.
Ans: Option (c) will give the most benefits because the use of good quality seeds is not only sufficient until the soil is properly irrigated, enriched with fertilisers and protected from biotic factors.
Page No. 146
Q1: Why should preventive measures and biological control methods be preferred for protecting crops?
Ans: Over-exposure to chemicals leads to environmental problems; hence, biological methods are preferred for protecting crops from pathogens, insects and rodents, along with increasing production. Since chemicals are harmful to plants and also to the animals that feed on them, bio-pesticides are used as a safe way of crop protection.
Q2: What factors may be responsible for the losses of grains during storage?
Ans: Biotic and Abiotic factors are responsible for the loss of grains during storage like
Rodents
Pests
Insects
Fungi
Bacteria
Sunlight
Flood
Rain
Temperature
Moisture
Page No. 147
Q1: Which method is commonly used for improving cattle breeds and why? How is cross-breeding useful in animals?
Ans: To improve the cattle breeds, we generally use the cross-breeding method. It is a process in which a cross is made between indigenous varieties of cattle by exotic breeds to get a crossbreed that is high-yielding. During cross-breeding, the desired characters taken into consideration are that the offspring should be high-yielding, should have early maturity and should be resistant to diseases and climatic conditions.
Page No. 148
Q1: Discuss the implications of the following statement. It is interesting to note that poultry is India’s most efficient converter of low-fibre foodstuff (which is unfit for human consumption) into highly nutritious animal protein food.
Ans: Poultry farming aims to raise domestic birds for egg and chicken meat purposes. These domestic birds feed on animal feeds which mainly consist of roughages for getting good quality feathers, eggs, chicken and nutrient-rich manure. For these reasons, it is said that “poultry is India’s most efficient converter of low-fibre foodstuff into highly nutritious animal protein food.”
Q2: What management practices are common in dairy and poultry farming?
Ans:
Well-designed Hygienic shelter for dairy animals and poultry birds.
Good quality, proper food and fodder are provided to dairy animals and poultry birds.
Importance for animal health by prevention and cure of diseases caused by bacteria, viruses, or fungi.
Sunlight-feasible and air,y ventilated shelter for animals.
Q3: What are the differences between broilers and layers and in their management? Ans:
Broilers
The poultry bird raised for meat purposes is called a broiler. Broilers feed on protein-rich, adequate-fat food. The level of vitamins A and K is kept high in poultry feeds.
Layers
The egg-laying poultry bird is called a layer. The housing, environmental and nutritional requirements of broilers vary from those of egg layers. Layers require proper lighting and enough space.
Page No. 150
Q1: How are fish obtained?
Ans: Fishes are obtained in two ways:
Capture fishing: Obtaining fish from natural resources.
Culture Fishery: Culturing of fish in freshwater ecosystems, like rivers, ponds and lakes, also including marine.
Q2: What is the advantage of composite fish culture?
Ans: The advantages of composite fish culture are:
In a single fish pond, a combination of 5 or 6 types of fish species can be cultured since they do not compete for food among themselves
Food resources can be completely utilised
Survival of the fish also increases
More yield
Q3: What are the desirable characters of the varieties suitable for honey production?
Ans: Desirable characters in varieties for honey production are:
The variety of bees should yield a large amount of honey.
The bees should stay for a longer period in the beehives.
The bees should not sting much.
Bees should be disease-resistant.
Honey Production
Q4: What is pasturage and how is it related to honey production?
Ans: Pasturage refers to the availability of flowers to the bees for easy accessibility for pollen collection and nectar. The kinds of flowers available will determine the taste of the honey; hence, Pasturage is the main reason for good-quality honey.
Page No. 151
Exercises
Q1: Explain any one method of crop production which ensures high yield.
Ans: Plant breeding is one of the methods adopted for high-yield plant breeding and is implemented to improve the varieties of crops by breeding plants. Plants from various places/areas are picked up with preferred traits, and then the process of hybridisation or cross-breeding is done among these diversities to get a crop/plant of anticipated characteristics.
Q2: Why are manure and fertilisers used in fields?
Ans: Manures and fertilisers are used to enrich the soil quality and improve the yield. They also help in controlling diseases. Manure and fertilisers replenish the soil by supplying nutrients to the soil. They are excellent sources of potassium, phosphorus and nitrogen, which assist in the healthy development of plants. Manures and fertilisers mainly improve the fertility of the soil.
Q3: What are the advantages of inter-cropping and crop rotation?
Ans: Inter-cropping
Checks pests and rodents, and hence decreases the chances of the spoiling of whole crops
Decreased chances of soil erosion
Reduced loss of crops with high-yield
Less water requirement
Crop rotation
Farmers can grow two or three crops annually
Pulses take nitrogen directly from the atmosphere and hence require a minimal amount of fertilisers
Both fruits and vegetables can be grown easily
Best use of land with a proper supply of nutrients
Q4: What is genetic manipulation? How is it useful in agricultural practices?
Ans:
Genetic manipulation is a process in which the transfer of genes takes place from one organism to another. Here, a gene of a particular character is introduced into the chromosome cell, resulting in a transgenic plant.
Example: Bt Cotton is a genetically modified crop that carries bacterial genes that protect this plant from insects. These are used in plants like brinjal, cabbage, rice, cauliflower, and maize crops to get protection from insects.
Q5: How do storage grain losses occur?
Ans: Storage grain losses occur due to various abiotic and biotic factors.
Abiotic factors
Humidity
Air
Temperature
Flood
Wind
Biotic factors
Insects
Rodents
Pesticides
Bacteria
Mites
Birds
Q6: How do good animal husbandry practices benefit farmers?
Ans: Good practice of animal husbandry benefits farmers in the following ways:
Yields in good-quality cattle
Better quality of milk production
Use in agriculture for carting, irrigation and tilling
Q7: What are the benefits of cattle farming?
Ans: The benefits of cattle farming are
Cattle are used for agricultural purposes
Generation of good-quality cattle
Milking and meat purpose
The skin of cattle is used for the leather and wool industry
Q8: For increasing production, what is common in poultry, fisheries, and beekeeping?
Ans: For increasing production, cross-breeding techniques are adopted in poultry, fisheries and bee-keeping. Along with these techniques, regular and proper maintenance methods are useful in improving production.
Q9: How do you differentiate between capture fishing, mariculture, and aquaculture?
Ans: Differences between capture fishing, mariculture, and aquaculture Fishing
Fish are obtained from natural resources, like ponds, canals, rivers,
Locating fish is easy and can be captured by using fishing nets.
Fishing
Mariculture
A method of marine fish culture in the open sea.
Fish can be located with the help of satellites and echo sounders. These can be caught by many kinds of fishing nets using fishing boats.
Mariculture
Aquaculture
Production of fish from freshwater and brackish water resources.
It can be located easily and caught using fishing nets.
Q1. How does the sound produced by a vibrating object in a medium reach your ear? Ans: When the object vibrates, it sets the neighbouring particles to vibrate. These particles exert force on other particles and pass on the energy to other parts of the medium. The particles do not get transported, but only the disturbance or energy is transferred. In this way, sound reaches our ears.
Q2. Explain how sound is produced by your school bell. Ans: When the hammer hits the gong of the bell, it starts vibrating. These vibrations set the particles of surrounding air vibrating. The disturbance travels in all directions and the sound propagates.
Q3. Why are sound waves called mechanical waves? Ans: Sound waves are produced by oscillations of particles of the medium. So they require a material medium for their propagation. Thus they are called mechanical waves.
Q4. Suppose you and your friends are on the moon. Will you be able to hear any sound produced by your friend? Ans: No, it is not possible to hear any sound on the moon. There is no medium such as air on the moon to carry sound waves. Sound cannot travel through a vacuum as it is a mechanical wave.
Page No. 132
Q1. Which wave property determines (a) loudness, (b) pitch? Ans: (a) Loudness is determined by the intensity or amplitude of sound waves. (b) Pitch is determined by the frequency of the sound wave.
Q2. Guess which sound has a higher pitch: guitar or car horn? Ans: Guitar, because it has a higher frequency.
Q3. What are the wavelength, frequency, time period and amplitude of a sound wave? Ans:Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength. It is devoted to A. Frequency: The number of complete waves produced per second is called the frequency of the wave. Time Period: The time taken to complete one complete vibration is called time period. Amplitude: The maximum displacement of the particles of the medium from their mean positions during the propagation of a wave is called the amplitude of the wave.
Q4. How are the wavelength and frequency of a sound wave related to its speed? Ans: Speed of sound = Frequency × Wavelength
Q5. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium. Ans: Given that, Frequency of sound wave = 220 Hz Speed of sound wave = 440 m/s Calculate wavelength. We know that Speed = Wavelength × Frequency v = λ ν 440 = Wavelength × 220 Wavelength = 440/220 Wavelength = 2 Therefore, the wavelength of the sound wave = 2 metres
Q6. A person is listening to a tone of 500 Hz, sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source? Ans: The time interval between successive compressions from the source is equal to the time period, and the time period is reciprocal to the frequency. Therefore, it can be calculated as follows: T= 1/F T= 1/500 T = 0.002 s
Page No. 133
Q1. Distinguish between loudness and intensity of sound. Ans:
Q2. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature? Ans: Sound travels fastest in solids due to their high elasticity, so out of given media sound travels fastest in iron.
Page No. 134
Q1. An echo is returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms-1? Ans: If d is the distance of the reflecting surface from the source and t is the time interval of echo-return, then 2d = vt ⇒ d = vt/2 Here, v – 342 ms-1, t = 3 s ∴
Page No. 135
Q1. Why are the ceilings of concert halls curved? Ans: The ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall.
Page No. 136
Q1. What is the audible range of the average human ear? Ans: Audible range for human ear = 20 Hz – 20 kHz
Q2. What is the range of frequencies associated with: (a) Infrasound (b)Ultrasound? Ans: (a)Infrasound: Sound waves between the frequencies 1 to 20 Hz. (b)Ultrasound: Sound waves of frequencies above 20,000 Hz.
Page No. 138
Q1. What is sound and how is it produced? Ans: A sound is a form of energy that produces the sensation of hearing. It is produced by oscillation/ vibration of particles of a material medium.
Q2. Describe with the help of a diagram, how compressions and rarefactions are produced in the air near a source of the sound. Ans:
Q3. Why is sound wave called a longitudinal wave? Ans: In sound waves, the vibration of the particles of the medium is along the direction of wave propagation, so sound waves in the air are longitudinal.
Q4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with Miens in a dark room? Ans: It is the quality of sound (or waveform) that helps us to identify the voice of our friend without seeing him.
Q5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why? Ans:The speed of light is 3 x 108 m/s while that of sound in air is only 330 m/s. We know that time = distance/speed.
Since sound travels slower than light, thunder is heard after the flash is seen.
Q6. A person has a hearing range from 20 Hz to 20 kHz. what are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms1. Ans: For sound waves, Speed = Wavelength × frequency v = λ × v Speed of sound wave in air = 344 m/s (a) For v = 20 Hz λ1 = v/v1 = 344/20 = 17.2 m (b) For v2 = 20,000 Hz λ2 = v/v2 = 344/20,000 = 0.0172 m Therefore, for human beings, the hearing wavelength is in the range of 0.0172 m to 17.2 m.
Q7. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child. Ans: Consider the length of the aluminium rod = d Speed of sound wave at 25° C, V Al = 6420 ms-1 Time taken to reach the other end is, T Al = d/ (V Al) = d/6420 Speed of sound in air, V air = 346 ms-1 Time taken by sound to each other end is, T air = d/ (V air) = d/346 Therefore, the ratio of time taken by sound in aluminium and air is, T air / t Al = 6420 / 346 = 18.55
Q8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute Ans: Frequency = (Number of oscillations) / Total time Number of oscillations = Frequency × Total time Given, ∵ 1 minute = 60 s Vibrations in 1 s = frequency = 100. Vibrations in 60 s = 60 x 100 = 6000 times The source vibrates 6000 times in a minute and produces a frequency of 100 Hz.
Q9. Does sound follow the same laws of reflection as light does? Explain. Ans: Yes, sound and light follow the same laws of reflection given below :
The angle of incidence at the point of incidence = The angle of reflection.
At the point of incidence the incident sound wave, the normal and the reflected sound wave lie in the same plane.
Q10. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day? Ans: An echo is heard when the time interval between the reflected sound and the original sound is at least 0.1 seconds. As the temperature increases, the speed of sound in a medium also increases. On a hotter day, the time interval between the reflected and original sound will decrease, and an echo is audible only if the time interval between the reflected sound and the original sound is greater than 0.1 s.
Q11. Give two practical applications of reflection of sound waves. Ans: Applications of reflection of sound waves are
to locate underwater hidden objects such as rocks and icebergs through SONAR.
to detect any undesired objects in the sky near airports and borders of the country.
Q12. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms–2 and speed of sound = 340 ms–1. Ans: Height (s) of tower = 500 m Velocity (v) of sound = 340 m s−1 Acceleration (g) due to gravity = 10 m s−1 Initial velocity (u) of the stone = 0 Time (t1) taken by the stone to fall to the tower base As per second equation of motion: s= ut1 + (½) g (t1)2 500 = 0 x t1 + (½) 10 (t1)2 (t1)2 = 100 t1 = 10 s Time (t2) taken by sound to reach the top from the tower base = 500/340 = 1.47 s. t = t1 + t2 t = 10 + 1.47 t = 11.47 s. The time for the splash to be heard is 11.47 seconds from the time the stone is dropped.
Page No. 139
Q13. A sound wave travels at a speed of 339 ms–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible? Ans: Given: Velocity of sound, v = 339 ms-1 Wavelength, λ = 1.5 cm – 0.015 m Frequency, v = = 22.6 kHz This sound is not audible to human beings as its frequency is higher than the audible range.
Q14. What is reverberation? How can it be reduced? Ans: The time from generation of sound until its loudness reduces to zero is called reverberation time. The process due to which the persistence of sound is caused is called reverberation. This is reduced in an auditorium using sound-absorbent materials and good absorbers of sound.
Q15. What is loudness of sound? What factors does it depend on? Ans: The loudness of sound is a measure of the response of sound to our ears. The loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of the human ear detecting this energy. The loudness of sound is measured in decibels (dB). As energy reaching the ear depends on the square of the amplitude, the loudness of sound depends on two factors: (i) Amplitude of sound waves and (ii) Sensitivity of ear.
Q16. How is ultrasound used for cleaning? Ans: Ultrasound is used to clean the hard-to-reach places such as spiral tubes, electronic components etc. The object to be cleaned is placed in the cleaning solution and ultrasonic waves are sent into it. The high frequency of ultrasound detaches the dust, grease and dirt from the object and it gets thoroughly cleaned.
Q17. Explain how defects in a metal block can be detected using ultrasound. Ans: Ultrasonic waves are allowed to pass through the metal. If the block is flawless, it will pass through it. If there is a crack or deformity, the wave gets reflected. The time taken by the wave to return is measured and helps to locate the flaw. If the wave is not reflected, it means the metal has no deformity.
Q1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (fig given). Let us take it that the force acts on the object through displacement. What is the work done in this case?
Ans: When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by: Work done = Force × Displacement W = F × S Given: Force exerted, F = 7 N Displacement, S = 8 m Therefore, work done, W = 7 × 8 = 56 Nm = 56 J
Page No. 116
Q1. When do we say that work is done? Ans: Work is done whenever the given conditions are satisfied:
A force acts on the body.
There is a displacement of the body caused by the applied force along the direction of the applied force.
For example:
If a bullock pulls a cart and it moves, work is done.
However, if a force is applied but there is no movement, no work is done.
Q2. Write an expression for the work done when a force is acting on an object in the direction of its displacement. Ans: When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression: Work done = Force × Displacement
Q3. Define 1 J of work. Ans: 1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.
Q4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field? Ans: Work done by the bullocks is given by the expression: Work was done = Force × Displacement W = F × d Given that: Applied force, F = 140 N Displacement, d = 15 m Therefore, W = 140 × 15 = 2100 J Hence, 2100 J of work is done in ploughing the length of the field.
Page No. 119
Q1. What is the kinetic energy of an object? Ans: Kinetic energy is a form of energy associated with the motion of an object. It is a crucial factor in determining the amount of work an object can do based on its movement. All moving objects possess kinetic energy, and this energy can be transformed into other forms to perform tasks.
The formula for kinetic energy is:
Ek = ½ mv²
Where:
Ek = kinetic energy
m = mass of the object
v = velocity of the object
Example:
A moving hammer uses its kinetic energy to drive a nail into a piece of wood.
Similarly, windmills harness the kinetic energy of moving air to generate electricity. Windmills harnessing the Kinetic energy
Q2. Write an expression for the kinetic energy of an object. Ans: If a body of mass m is moving with a velocity v, then its kinetic energy is given by the expression: Kinetic energy, Ek = 1/2 mv2 Its SI unit is Joule (J).
Q3. The kinetic energy of an object of mass, m moving with a velocity of 5 ms–1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times? Ans: Expression for kinetic energy is Ek = 1/2 mv2 m = Mass of the object v = Velocity of the object = 5 ms−1 Given that kinetic energy, Ek= 25 J
If the velocity of an object is doubled, then v = 5 × 2 = 10 m s−1. Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225 J.
Page No. 123
Q1. What is power? Ans: Power is the rate of doing work or the rate of transfer of energy. If ‘W’ is the amount of work done in time ‘t’, then power is given by the expression
It is expressed in Watt (W).
Q2. Define 1 watt of power. Ans: A body is said to have a power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e., 1 W = 1 J / 1 s
For larger rates of energy transfer, we use kilowatts (kW):
1 kW = 1000 W
1 kW = 1000 J/s
Q3. A lamp consumes 1000J of electrical energy in 10s. What is its power? Ans: Power is given by the expression:
Work done = Energy consumed by the lamp = 1000 J Time = 10 s = 100 W
Q4. Define average power. Ans: Average power is obtained by dividing the total amount of work done in the total time taken to do this work.
Page No. 124
Exercises
Q1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
Suma is swimming in a pond.
A donkey is carrying a load on its back.
A windmill is lifting water from a well.
A green plant is carrying out photosynthesis.
An engine is pulling a train.
Food grains are getting dried in the sun.
A sailboat is moving due to wind energy.
Ans: (i) Suma is swimming in a pond: In this case, Suma pushes the water in the backward direction. It is known as action. However, due to the reaction, the water pushes the person in the forward direction. In this situation, the force and displacement are in the same direction. The work is done by Suma and the work done by the force is positive.
(ii) A donkey is carrying a load on its back: If the displacement of an object is perpendicular to the force acting on it, the work done by the force on the object is zero.
In the given situation, the force of gravity on the load is in the downward direction, whereas the displacement is in the horizontal direction, i.e., the force and the displacement are perpendicular to each other. There is no displacement in the direction of the force of gravity, and therefore, the work done is zero. (iii) A windmill is lifting water from a well: In this situation, the object (bucket of water from a well) moves upwards. The force exerted by the windmill is in the direction of displacement. Therefore, work is done. This work is positive as the force and the displacement are in the same direction. (iv) A green plant is carrying out photosynthesis: In this case, no work is done because photosynthesis is a chemical process that converts light energy into chemical energy, rather than involving any physical movement or mechanical work. (v) An engine is pulling a train: In this situation, an engine is pulling a train parallel to the ground. The force exerted by the engine is in the direction of displacement of the train. Thus, the force and the displacement are in the same direction. Therefore, work is done. This work done is positive. (vi) Foodgrains are getting dried in the sun: No work is done in this case. (vii) A sailboat moving due to wind energy: The force exerted by wind on the sail moves the boat in the direction of force, hence, positive work is done by wind energy.
Page No. 125
Q2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object? Ans: Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions/heights of the object, which is zero. Work done by gravity is given by the expression: W = mgh Where: h = Vertical displacement = 0 W = mg × 0 = 0 J Therefore, the work done by gravity on the given object is zero joule.
Q3. A battery lights a bulb. Describe the energy changes involved in the process. Ans:
When a bulb is connected to a battery, then the chemical energy of the battery is transferred into electrical energy.
When the bulb receives this electrical energy, then it converts it into light and heat energy.
Hence, the transformation of energy in the given situation can be shown as: Chemical Energy(battery) → Electrical Energy (battery to bulb) → Light Energy (from the bulb)+ Heat Energy(from the bulb).
Q4. Certain force acting on a 20kg mass changes its velocity from 5ms−1 to 2m s−1. Calculate the work done by the force. Ans: Kinetic energy is given by the expression: (Ek)v = 1/2 mv2 Where: Ek = Kinetic energy of the object moving with a velocity, v m = Mass of the object (i) Kinetic energy when the object was moving with a velocity 5ms−1
Kinetic energy when the object was moving with a velocity 2ms−1
Work done by force is equal to the change in kinetic energy. Therefore, work done by force = (Ek)2 – (Ek)5 = 40 − 250 = −210 J The negative sign indicates that the force is acting in the direction opposite to the motion of the object.
Q5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. Ans: Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression: W = mgh Where: Vertical displacement, h = 0 ∴W = mg × 0 = 0 Hence, the work done by gravity on the body is zero.
Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why? Ans:
No. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then its potential energy changes into kinetic energy progressively.
A decrease in the potential energy is equal to an increase in the kinetic energy of the body.
During the process, total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.
Q7. What are the various energy transformations that occur when you are riding a bicycle? Ans:
While riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle.
Heat energy heats the rider’s body. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as: During the transformation, the total energy remains conserved.
Q8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going? Ans: When exerting force on a large, immovable rock, our muscles’ energy is not transferred to the rock. Instead, this energy is converted into heat, leading to an increase in our body temperature, without any loss of energy.
Your effort results in increased body heat.
No movement occurs in the rock, so no work is done on it.
The energy you expend is not wasted; it simply changes form.
Q9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules? Ans: 1 unit of energy is equal to 1 kilowatt-hour (kWh). 1 unit = 1 kWh 1 kWh = 3.6 × 106 J Therefore, 250 units of energy = 250 × 3.6 × 106 = 9 × 108 J
Q10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. Ans: Gravitational potential energy is given by the expression: W = mgh Where: h = Vertical displacement = 5 m m = Mass of the object = 40 kg g = Acceleration due to gravity = 9.8 m s−2 ∴ W = 40 × 5 × 9.8 = 1960 J. At half-way down, the potential energy of the object will be 1960/2 = 980 J. At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.
Q11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer. Ans: Work is done whenever the given two conditions are satisfied: (i) A force acts on the body. (ii) There is a displacement of the body by the application of force in or opposite to the direction of force. If the direction of force is perpendicular to displacement, then the work done is zero. When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the Earth is zero.
Q12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher. Ans: Yes, there can be displacement without a force acting on an object.
If an object moves with a constant velocity, the net force acting on it is zero.
Despite the absence of a net force, the object still experiences displacement along its path.
Therefore, it is possible for an object to be displaced even when no force is applied.
Q13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer. Ans: Work is done whenever the given two conditions are satisfied: (i) A force acts on the body. (ii) There is a displacement of the body by the application of force in or opposite to the direction of force. When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Although the force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.
Q14. An electric heater is rated 1500 W. How much energy does it use in 10 hours? Ans: Energy consumed by an electric heater can be obtained with the help of the expression:
Where: Power rating of the heater, P = 1500 W = 1.5 kW Time for which the heater has operated, T = 10 h Work done = Energy consumed by the heater Therefore, energy consumed = Power × Time = 1.5 × 10 = 15 kWh Hence, the energy consumed by the heater in 10h is 15 kWh.
Q15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy? Ans: The law of conservation of energy states that energy can neither be created nor destroyed. It can only be converted from one form to another. Consider the case of an oscillating pendulum:
In a pendulum, as it swings, energy alternates between kinetic energy (KE) and potential energy (PE):
At extreme positions (A or B): The pendulum rises to a height h, where its KE is zero, and it has maximum PE.
At the mean position (P): The pendulum is at its lowest point with maximum speed. Here, PE is zero, and it has maximum KE.
This energy exchange continues during oscillation. However, due to air resistance, the pendulum loses KE, which is converted to heat in the surroundings. Eventually, the pendulum stops. The law of conservation of energy holds because the total energy (pendulum + surroundings) remains constant, even though the pendulum’s energy decreases.
Page No. 126
Q16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest? Ans: Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression: In order to bring it to rest, its velocity has to be reduced to zero, and in order to accomplish that, the kinetic energy has to be drained off and sent somewhere else. An external force has to absorb energy from the object, i.e. do negative work on it, equal to its kinetic energy, or – 1/2 mv2.
Q17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h? Ans: Kinetic energy, Ek= 1/2 mv2 Where: Mass of car, m = 1500 kg Velocity of car, v = 60 km/h = 60 x (5/18) m/s
Hence, 20.8 × 104 J of work is required to stop the car.
Q18. In each of the following a force, ‘F’ is acting on an object of mass, ‘m’. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Ans: Work is done whenever the given two conditions are satisfied: (i) A force acts on the body. (ii) There is a displacement of the body by the application of force in or opposite to the direction of force. Case I:
In this case, the direction of force acting on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero. Case II:
In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive. Case III:
In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by force on the block will be negative.
Q19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why? Ans: Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other, i.e., the net force acting on the object is zero. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.
Q20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each. Ans: Energy consumed by an electric device can be obtained with the help of the expression for power: P = W / T Where: Power rating of the device, P = 500 W = 0.50 kW Time for which the device runs, T = 10 h Work done = Energy consumed by the device Therefore, energy consumed = Power × Time = 0.50 × 10 = 5 kWh Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20 kWh = 20 Units.
Q21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy? Ans:
When an object falls freely towards the ground, its potential energy decreases and kinetic energy increases.
As the object touches the ground, all its potential energy gets converted into kinetic energy.
As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. It can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.
Page No. 102Q1.State the universal law of gravitation. Ans: Everybody in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.Universal Law of GravitationQ2.Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth. Ans: Consider F as the force of attraction between an object on the surface of the earth and the earth. Let the mass of earth = M Mass of object = m. If the radius of the object is comparatively negligible, r ≈ radius of earth = R F = GMm/R2
Page No. 104
Q1. What do you mean by free fall? Ans: When an object falls with a constant acceleration, under the influence of the force of gravitation of the Earth, the object is said to have a free fall.
Q2. What do you mean by acceleration due to gravity? Ans: During the course of its free fall, a body accelerates due to the force of gravity acting on it. This acceleration is known as acceleration due to gravity.
Page No. 106
Q1. What are the differences between the mass of an object and its weight? Ans:
Q2. Why is the weight of an object on the moon 1/6th its weight on the Earth? Ans: Mass of the moon (M) = 7.4 × 1022 kg The radius of the moon (R) = 1.74 × 106 m Gravitational constant (G) = 6.7 × 1011 Nm2kg2 We know that acceleration due to gravity(g) = GM/R2 Also, weight, W = mg W = GMm/R2 Let the mass of the object be m, let its weight of the moon be Mm andits radius be Rm By applying the universal law of gravitation, the weight of the object on the moon will be:
(1)
Let the weight of the same object on the earth be We. The mass of the earth is M and its radius is R.
(2)
Substituting the values in equations (1) and (2) we get:
Wm= 2.431 x 1010G x m
and We = 1.474 x 1011 G x m
On dividing Wm and We we get:
∴ The weight of the object on the moonits weight on Earth
Page No. 109
Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string? Ans:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large.
This is because the pressure is inversely proportional to the surface area on which the force acts.
The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.
Q2. What do you mean by buoyancy? Ans:
The upward force exerted by a liquid on an object immersed in it is known as buoyancy.
When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water.
Q3. Why does an object float or sink when placed on the surface of water? Ans:
If the density of an object is more than the density of the liquid, then it sinks in the liquid.
This is because the buoyant force acting on the object is less than the force of gravity.
On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid.
This is because the buoyant force acting on the object is greater than the force of gravity.
Page No. 110
Q1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? Ans: Mass is always a constant quantity. Therefore, it cannot be more or less than 42 kg.
Q2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why? Ans: The correct answer is cotton bag is heavier than an iron bar. The reason is:
True weight = (apparent weight + up thrust)
The density of the cotton bag is less than that of the iron bar, so the volume of the cotton bag is greater than the iron bar. So the cotton bag experiences more upthrust due to the presence of air.
Therefore in the presence of air, the true weight of a cotton bag is greater than the true weight of an iron bar.
Page No. 111
Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half? Ans: The force of gravitation between two objects of masses m1 and m2 at separation r is given by: F = (G m1m2)/r2 When the distance between them is reduced to half, the gravitational force. F’ = (G m1m2)/r2 = (G m1m2)/ (r/2)2 = (4 G m1m2)/(r2) = 4F The gravitational force becomes four times.
Q2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object? Ans: The acceleration due to gravity of a freely falling body is independent of the mass of the falling body. Thus, both heavy and light objects fall with the same acceleration.
Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m.) Ans: Here, the mass of the earth, M = 6 x 1024 kg Mass of the object, m = 1 kg, radius of the earth, R = 6.4 x 106 m, G = 6.7 x 10-11 Nm2kg-2 The magnitude of the gravitational force between the earth and the object is given by: The force of gravitation between them,
Q4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller force or the same as the force with which the moon attracts the earth? Why? Ans:
Earth attracts the moon with an equal and opposite force as the moon attracts the Earth.
This is in accordance with Newton’s law of gravitation according to which gravitational force is the mutual force of attraction which is also equal as per to the third law of motion.
Q5. If the moon attracts the Earth, why does the earth not move towards the moon? Ans:
The moon revolves around the earth because the gravitational force of the earth on the moon provides the necessary centripetal force for circular motion.
The moon also attracts the earth, due to which the earth also revolves around the moon in a circular orbit, but due to the very large mass of the earth, the motion of the earth is not observed.
It may be kept in mind that there is no linear motion of Earth towards the moon or the moon towards Earth. The gravitational force between them keeps on changing the direction of motion of the moon or earth in a circular orbit.
Q6. What happens to the force between two objects, if (a) the mass of one object is doubled? (b) the distance between the objects is doubled and tripled? (c) the masses of both objects are doubled? Ans: Let masses = m1 and m2, the distance between masses = r. According to the universal law of gravitation F ∝ m1m2 and F ∝ 1/r2. (a) If one mass, say m1 is doubled, and F ∝ m1, then F gets doubled. (b) If distance is doubled and F ∝ 1/r2, force becomes one-fourth, i.e. 1/(2)2. If the distance is tripled and F ∝ 1/r2, force becomes one-ninth, i.e. 1/(3)2. (c) If both masses are doubled and F ∝ m1m2, force becomes 4 times, i.e. (2m1)(2m2) = 4m1m2
Q7. What is the importance of universal law of gravitation? Ans: The universal law of gravitation has successfully explained the phenomena that were earlier considered to be separate: (a) It explains the force which holds us on earth. (b) It explains the motion of planets around the sun. (c) It explains the motion of the moon around the Earth. (d) It explains the occurrence of tides in the ocean.
Q8. What is the acceleration of free fall? Ans: When a body falls freely and accelerates at every point of its motion due to gravitational force alone. This is called the acceleration of free fall. This acceleration is known as the acceleration due to gravity on the earth’s surface. It’s denoted by ‘g’, and its value is 9.8m/s2, and it’s constant for all objects close to the earth’s surface (irrespective of their masses).
Q9. What do we call the gravitational force between the earth and an object? Ans: Gravity is the name of the gravitational force between the earth and the object.
Q10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.] Ans: Earth is not a perfect sphere. It is flattened at the poles. Thus, the value of ‘g’ is greater at the poles than at the equator (as g ∝ 1/r2). This means the weight of gold bought at the poles becomes lesser at the equator, and the friend does not agree with the weight.
Q:11. Why will a sheet of paper fall slower than one that is crumpled into a ball? Ans:
A sheet crumpled into a ball has a small surface area as compared to that of an unfolded sheet. Therefore, the unfolded sheet will experience more friction due to air as compared to the crumpled ball in spite of the same force of gravity acting upon it.
The larger friction of air slows down the unfolded sheet so it will fall slower than the sheet crumpled into a ball.
Q12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth? Ans: Here, mass of object, m = 10 kg, g on the earth’s surface = 9.8 ms-2 Weight of the object on the earth’s surface = mg = 10 kg x 9.8 ms-2 = 98.0 kg ms-2 = 98 N The weight of an object on the moon is 1/6th that of the object on Earth, so the weight of the object on the moon: = 1/6 x 98 N= 16.34 N.
Page No. 112
Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (a) the maximum height to which it rises. (b) the total time it takes to return to the surface of the earth. Ans: (a) Initial velocity, u = 49 ms-1, Final velocity, v = 0 Maximum height, H = ? Acceleration due to gravity, g = + 9.8 ms-2 (downward)
As object is moving upward, so a = -gFrom relation v2 = u2 + 2as , we have a = -g , s = H v2 = u2 – 2gH
⇒ H = 122.5 m (b) When the ball returns to the surface of the earth, displacement, s = 0 ∴ Relation, s = ut + at2 gives s = ut -gt2 0 = (49 ms-1) t -x 9.8 t2 ⇒ t (49 -4.9 t) = 0 t = 0 or 49 – 4.9 t = 0 ⇒ t = As t ≠ 0, so t = 10s The maximum height is 122.5 m and the total time is 10 s.
Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. Ans: Initial velocity, u = 0, height, h = 19.6 m and g = 9.8 ms-2 Final velocity,
= 19.6 ms-1.
Q15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone? Ans: Initial velocity, u = 40 ms-1 g = -10 ms-2 (upward motion) Final velocity, v = 0 During upward motion, g = -10 ms-2
Net displacement on returning back = zero Total distance = 80 m + 80 m = 160 m
Q16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m. Ans: Mass of earth, Me = 6 x 1024 kg Mass of sun Ms = 2 x 1030 kg and Distance, r = 1.5 x 1011 m Gravitational constant, G = 6.67 x 10-11 Nm2/kg2 Force,
= 3.56 x 1022 N
Q17. A stone is allowed to fall from the top of a tower 100 m high and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. Ans: u1 = 0 Let the stones meet at a height h above the ground. Stone 1 covers a distance (100 – h); g = 10 m/s2. (100 – h) = 0 + ⇒ 100 – h = 5 t2 …(i) u2 = 25 m/s Stone 2 covers a distance h g = – 10m/s2 h = u2t – ⇒ h = 25 t – 5 t2 …(ii) From (i) and (ii), we get 100 – 25 t + 5 t2 = 5 t2 100 – 25 t = 0 or t = 4 s (Stones meet 4 s after they are thrown) h = 25(4) – 5(4)2 = 100 – 80 = 20 m above the ground.
Q18. A ball thrown up vertically returns to the thrower after 6s. Find (a) the velocity with which it was thrown up. (b) the maximum height it reaches. (c) its position after 4s. Ans:
Time of ascend = Time of descending
⇒ Time taken to reach the top, t = 3 s. ⇒ Final velocity, v = 0, g = -9.8 ms-2. (a) Initial velocity, u = v – gt = 0 – (-9.8) (3) = 29.4 ms-1 (b) Maximum height reached, h = (v2 – u2)/g = (29.4)2/(2 x 9.8) = 44.1 m (c) After 4s, the ball has started falling and has fallen by some distance h’ for 1s. Here, initial velocity u’ = 0, t = 1s g = + 9.8 ms-1 h’ = u’t + = 0 + The ball is at a height, (44.1 – 4.9) = 39.2 m above the ground.
Q19. In what direction does the buoyant force on an object immersed in a liquid act? Ans: An object immersed in a liquid experiences buoyant force in the upward direction.
Q20. Why does a block of plastic released under water come up to the surface of water? Ans:
Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards.
If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within the water.
Due to this reason, a block of plastic released under water comes up to the surface of the water.
Q21. The volume of 50g of a substance is 20 cm3. If the density of water is 1g cm-3, will the substance float or sink? Ans:
If the density of an object is more than the density of a liquid, then it sinks into the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Here, the density of the substance =
The density of the substance is more than the density of water (1 g cm−3). Hence, the substance will sink in water.
Q22. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3? What will be the mass of the water displaced by this packet? Ans:
Density of the 500 g sealed packet
The density of the substance is more than the density of water (). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 cm3 of water has a mass of 350 g.
Q1.Which of the following has more inertia? (a) A rubber ball and a stone of the same size? (b) A bicycle and a train? (c) A five rupees coin and one-rupee coin? Ans: Inertia is the property of an object to resist a change in its state of motion. More mass = More inertia (a) A stone has more mass than a rubber ball of the same size. Therefore, the stone has more inertia. (b) A train has much more mass than a bicycle. Therefore, the train has more inertia. (c) A five-rupee coin is heavier than a one-rupee coin. Therefore, the five-rupee coin has more inertia.
Q2.In the following example, try to identify the number of times the velocity of the ball changes. “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case. Ans: The velocity of the ball changes four times in the following ways: (a) Player 1 changes the velocity by kicking it. (b) Player 2 changes the velocity by kicking it to the goal. (c) The goalkeeper changes the velocity by, collecting the ball. (d) The goalkeeper changes the velocity by kicking the ball.
Q3.Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Ans: The leaves of the tree have inertia of rest. When the branch is shaken, it moves, but the leaves tend to be in a state of rest due to their inertia. Thus, they get detached from the tree.
Q4.Why do you fall in the forward direction when a moving bus apply brakes to stop and fall back when it accelerates from rest? Ans:
In a running bus, our speed is equal to the speed of the bus. As a moving bus brakes to a stop, the lower part of our body being in contact with the bus comes to rest, but the upper part due to inertia of motion remains in the state of motion. The bus stops suddenly, Passenger jerks forward
Hence, we fall in the forward direction. When the bus accelerates from rest, the feet come into motion while the upper part of the body remains at rest due to the inertia of rest; hence, we fall backwards.
Page No. 97
Exercises
Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. Ans:
Yes, it is possible for the object to be travelling with a non-zero velocity if it experiences a net zero external unbalanced force.
This is due to the inertia of motion. If the body is initially moving with some velocity on a smooth surface, then it will continue to move with the same velocity, though the net external force acting on the body is zero. Example: When we stop pedalling a moving bicycle, the bicycle begins to slow down and finally comes to rest. This is again because of the friction forces acting opposite to the direction of motion. The force of friction opposes the motion of the bicycle. If there were no unbalanced force of friction and no air resistance, a moving bicycle would go on moving forever.
Q2.When a carpet is beaten with a stick, dust comes out of it. Explain. Ans: When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to the inertia of rest, therefore, dust comes out of it.
Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Ans: The luggage kept on the roof possesses inertia. As the bus moves, brakes to stop or takes a turn, the luggage might fall off. Thus, it should be tied with a rope.
Q4.A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. Ans: When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c). If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.
Q5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.) Ans: Initial velocity of truck (u) = 0 Time (t) = 20 s Distance covered (S) = 400 m Acceleration (a) = ? Mass of truck (m) 7 tonne = 7000 kg Force on truck We know ; (F) = ? We know: S= ut 1/2 at2 400 = 0 × 20 1/2 × a × (20)2 400 = 200a a = 2ms-2 Force on truck (F) = ma = 7000 x 2 = 14000 N
Q6.A stone of 1 kg is thrown with a velocity of 20 ms–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Ans: Mass, m = 1kg; Initial velocity, u = 20 ms-1; Final velocity, v = 0; distance travelled, s = 50 m Acceleration,
Force exerted = ma = 1 kg x (- 4) ms-2 = – 4N Friction = 4 N against the direction of motion.
Q7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train Ans:(a) Given, the force exerted by the train (F) = 40,000 N The force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction) Therefore, the net accelerating force = the sum of all forces
= 40,000 N + (-5000 N)
= 35,000 N (b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg The total mass of the train is 18000 kg. As per the second law of motion, F = ma (or: a = F/m) Therefore, acceleration of the train = (net accelerating force) / (total mass of the train) = 35,000/18,000 = 1.94 ms-2 The acceleration of the train is 1.94 ms-2
Q8.An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of1.7 ms–2? Ans: Here, the mass of the automobile, m = 1,500 kg Acceleration, a = 1.7 ms-2 Force between the vehicle and road (F) can be calculated as F = ma = 1500 kg x ( -1.7 ms-2) = -2,550 kg ms-2 = -2,550 N
Negative sign indicates that the force is in a direction opposite to the motion of the vehicle.
Page No. 98
Q9.What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2 (b) mv2 (c) 1/2mv2 (d) mv Ans: (d) mass x velocity The momentum of an object is defined as the product of its mass m and velocity v. Momentum = mass x velocity Hence, the correct answer is mv, i.e., option (d).
Q10.Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet? Ans: Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of the opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.
Q11.According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal force cancel each other Comment on this logic and explain why the truck does not move. Ans: The logic given by the student is wrong because the action and reaction forces do not act on the same body, but act on different bodies. The reason for not moving the truck is as follows:
The truck is massive, so it has a very large inertia. Moreover, it is parked on the ground; there is a frictional force between the truck and the ground.
The force exerted by us on the truck is insufficient to overcome the force of friction; so these are balanced forces on the truck; (force exerted by us plus force of friction in opposite direction); that is, the net force is zero; and hence, the truck does not move.
Q12.A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. Ans: Mass of ball (m) = 200 g = 0.2 kg Initial velocity of ball (u1) = 10 ms–1 Final velocity of ball (u2) = 5 ms–1 (Negative sign denotes that the ball is moving in the opposite direction) Initial momentum of ball = mu1 = 0.2 × 10 = 2 Ns Final momentum of ball = mu2 = 0.2 × (- 5) = -1 Ns Change in momentum = Final momentum – Initial momentum = -1 – 2 Ns = -3 Ns
Negative sign denotes that change in momentum is in the direction opposite to the direction of initial momentum of the ball.
Q13.A bullet of mass 10g travelling horizontally with a velocity of 150 ms–1 strikes a stationary wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet. Ans: Mass of bullet (m)= 10 g = 0.01 kg The initial velocity of a bullet (u) =150 ms–1 The final velocity of a bullet (v) = 0 Time (t)= 0.03 secs Acceleration on the bullet (a) = ? Force acting on wooden block (F) = ? Distance penetrated by the bullet (s) = ? We know: v = u + at 0 = 150 + (a ×0.3) a × 0.03 = -150 a = -5000 ms-2 (Negative sign indicates that the velocity of the bullet is decreasing.)
We know:
S = ut + 0.5at2 s = 150 × 0.03 + (-5000) × (0.03)2 = 4.5 + 2.25 = 6.75m We know: F = ma Force acting on bullet (F) = ma = 0.01 × (-5000) = -50N
Negative sign denotes that wooden block exerts force in the direction, opposite to the direction of motion of the bullet.
Q14.An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1 collides with and sticks to a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. Ans: For object: m1 = 1 kg u1 = 10 ms1 For wooden block: m2 = 5 kg u2 = 0 Momentum just before collision: = m1u1 + m2u2 = 1 × 10 + 5 × 0 = 10 kg ms–1 Mass after collision = (m1 + m2) = 1 + 5 = 6 kg Let velocity after collision = v Momentum after collision = 6 × v Using the law of conservation of momentum: Momentum after collision = Momentum before collision 6 × v = 10 v = 1.67 ms–1
Q15.An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. Ans: Mass of object (m) = 100 kg Initial velocity (u) = 5 ms–1 Final velocity (v) = 8 ms–1 Time (t) = 6 s Initial momentum (Pinitial) = mu = 100 × 5 = 500 Ns Final momentum (Pfinal) = mv = 100 × 8 = 800 Ns Force exerted on the object (F) = [(mv – mu) / t] = (800 – 500)/6 = 300/6 =50N
Q16.Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. Ans: As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass). Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high. Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal to the conservation of momentum. The velocity of an insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of a motorcar is very insignificant because its mass is very large compared to the insect. Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.
Q17.How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm and does not rebound? Take its downward acceleration to be 10ms–2. Ans: Height, h = 80 cm = 0.8 m Mass, m = 10 kg ; Initial velocity, u = 0 Acceleration, a = 10 ms-2 Final velocity, v =
Momentum transferred = mv = 10 kg x 4 ms-1 = 10 kg x 4 ms-1 = 40 kgms-1
Page No. 99
Q1.The following is the distance-time table of an object in motion:
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing or zero? (b) What do you infer about the forces acting on the object? Ans: (a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.
(b) Force, F = ma ⇒ F ∝ a. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well
Q2.Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms–2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.) Ans: Mass of car, m = 1200 kg Let each person apply force F on the motor car. Thus, the force ‘2F applied by 2 persons balances the frictional force due to the ground force applied by three persons = 3F Effective force = 3F – friction = 3F – 2F = F Thus force F produces an acceleration of 0.2 ms-2. F = ma = 1200 x 0.2 = 240 N Thus, force applied by each person = 240 N
Q3. A hammer of mass 500 g, moving at 50 ms–1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? Ans: Mass of hammer, M = 500 g = 0.5 kg The initial velocity of the hammer, u = 50 ms-1 The final velocity of the hammer, v = 0 Time, t = 0.01 s = – 5000 ms-2 The force applied by the nail, F = ma = (0.5) x (-5000 ms-2) = – 2500 N (opposite to the motion of the hammer)
Q4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required. Ans: Mass of motor car, m = 1200 kg Initial velocity of car, u = 90 km/h
Final velocity of car, v = 18 km/h
Time, t = 4s Acceleration, Momentum change = mv – mu or m(v – u) = 1200 (5 – 25) = – 24000 kg ms-1 Magnitude of force, F = ma = 1200 x (- 5) = – 6000 N Acceleration, momentum change and force are opposing the motion of the motorcar, as indicated by a negative sign.
Q1. An object has moved through a distance. Can it have zero displacements? If yes, support your answer with an example.
Ans: Yes, an object can have zero displacements even if it has moved through a distance.
Example: If an athlete runs around a circular path of radius ‘r’ and comes back to the initial point, then distance covered = 2πr, displacement = zero.
Q2. A farmer moves along the boundary of a square field of side 10 m in the 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Ans: Given,
Side of the square field = 10 m
Perimeter of the square = 4 × 10 = 40 m
Time to complete one round = 40 s
Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s
Speed of the farmer = Perimeter / Time = 40 / 40 = 1 m/s
Distance covered in 140 s = 1 × 140 = 140 m
Number of complete rounds = Total distance / Perimeter = 140 / 40 = 3.5
After 3 complete rounds (120 m, 120 s), the farmer is back at the starting point (e.g., point A at (0,0)). In the remaining 0.5 round (20 m, 20 s), the farmer moves halfway around the square. Starting at A (0,0), the path is:
A to B (10,0): 10 m
B to C (10,10): 10 m
After 0.5 round, the farmer is at C (10,10). The displacement is the straight-line distance from A (0,0) to C (10,10):
Q3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Ans:
(a)False; Displacement can be zero if an object returns to its initial position after moving. For example, an athlete running around a circular track and returning to the starting point has zero displacement.
(b)False; The magnitude of displacement is always less than or equal to the distance travelled. Displacement equals distance only when the motion is in a straight line without reversing direction. For example, in circular motion, displacement is zero when returning to the start, while distance is non-zero.
Page No. 76
Q1. Distinguish between speed and velocity.
Ans:
Q2. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Ans: Average speed measures the total distance covered over a specific time period, while average velocity refers to the total displacement during that same time. The magnitudes of average speed and average velocity will be equal when the total distance traveled matches the displacement.
Q3.What does the odometer of an automobile measure?
Ans: An odometer, also known as an odograph, is a device that calculates the distance an automobile has travelled by measuring the circumference of the wheel as it rotates.
Q4.What does the path of an object look like when it is in uniform motion?
Ans: The path of the object will be a straight line at the instant of measurement when it is in uniform motion.
Q5.During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is 3 × 108 ms–1.
Ans: Speed of signal = 3 × 108 ms–1
Time in which signal reaches ground = 5 min = 5 × 60 = 300 s
Distance of spaceship from the ground level = speed × time = 3 × 108 × 300 = 9 × 1010 mPage No. 77
Q1. When will you say a body is in
(i) Uniform acceleration (ii) Non-uniform acceleration?
Ans:
(i) If an object travels in a straight lineand its velocity increases or decreases by equalamounts in equal intervals of time, then the body is said to be in uniform acceleration.
Example: The motion of a freely falling body.
(ii) If an object travels in a straight line and its velocity changes by unequal amounts inequal intervals of time, then the body is said to be in non-uniform acceleration.
Example: If a car is travelling along a straight road and passes through a crowd, suffers an unequal change in velocity, in equal intervals of time.
Q2. A bus decreases its speed from 80 km h−1 to 60 km h−1 in 5 s. Find the acceleration of the bus.
Ans:
Initial velocity, u=80u=80km/h
Final velocity, v=60v=60km/h
Time, t=5t=5s
Solution:
Convert velocities from km/h to m/s:
Use the first equation of motion: v=u+atv=u+at
16.67=22.22+a⋅5
a⋅5=16.67−22.
a⋅5=−5.55
The acceleration of the bus is −1.11m/s2 (negative sign indicates deceleration or retardation).
Q3. A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h-1 in 10 minutes. Find its acceleration.
Ans: Given,
Initial velocity, u = 0 km /h
Final velocity, v = 40 km/h=40 × ( 5/ 18) = 11.11 m/s
Time, t = 10 min = 10 × 60 = 600 sec
Acceleration, a = ?
Consider the formula, v = u + at
⇒ 11.11 = 0 + a × 600
⇒ 11.11 = 600 a
⇒ a = 11.11/600 = 0.0185 m/s2
Page No. 81
Q1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Ans:
When the motion is uniform, the distance-time graph is a straight line with a slope.
When the motion is non-uniform, the distance-time graph is not a straight line. It can be any curve.
Q2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Ans: If the distance-time graph is a straight line parallel to the time axis, the body is at rest.
Q3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Ans: If the speed-time graph is a straight line parallel to the time axis, the object is moving uniformly.
Q4.What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans: The area beneath the velocity-time graph corresponds to the area of the rectangle OABC, calculated as OA multiplied by OC. Here, OA represents the object’s velocity, and OC indicates time. Thus, the shaded area can be expressed as:
The area under the velocity-time graph = velocity × time.
By substituting the value of velocity as displacement divided by time into this equation, we find that the area under the velocity-time graph represents the total displacement of the object.
Page No. 82
Q1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired, (b) the distance travelled.
Ans: Given,
Initial velocity, u = 0 ms-1
Acceleration, a = 0.1 ms-2
Time, t = 2 min = 120 s
(a) Speed, v = u + at = 0 + 0.1 x 120 = 12 ms-1
(b) Distance, s = 720 m
The speed acquired is 12 ms-1 and the total distance travelled is 720 m.
Q2.A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of 0.5 ms–2. Find how far the train will go before it is brought to rest.
Ans: Given theinitial speed of the train, u= 90 km/h = 25 m/s
Final speed of the train, v = 0 m/s (finally the train comes to rest)
Acceleration = – 0.5 m s-2
According to the third equation of motion:
v2 = u2 + 2as (where s is the distance covered by the train)
⇒ (0)2 = (25)2 + 2 (- 0.5) s
⇒
The train will cover a distance of 625 m at an acceleration of -0.5ms-2 before it comes to rest.
Q3. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s–2. What will be its velocity 3 s after the start?
Ans: Initial Velocity of trolley, u = 0 cms-1; Acceleration, a = 2 cms-2; Time, t = 3 s
We know that final velocity, v = u + at = 0 + 2 x 3 cms-1
Therefore, The velocity of the train after 3 seconds = 6 cms-1
Page No. 83
Q4.A racing car has a uniform acceleration of 4 ms-2. What distance will it cover in 10 s after the start?
Ans: Initial Velocity of the car, u = 0 ms-1; Acceleration, a = 4 m s-2; Time, t = 10 s
We know Distance, s = ut + (1/2) at2
Therefore, Distance covered by car in 10 second = 0 × 10 + (1/2) × 4 × 102 = (1/2) × 400 = 200 m
Q5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s-1. If the acceleration of the stone during its motion is 10 m/s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans: Given,
The initial velocity of stone, u = 5 ms-1
Downward or negative acceleration, a = 10 ms-2
We know that: 2as = v2– u2
⇒ 0 = (5)2 + 2 x(-10) x s
⇒ 0 = 25 – 20s
⇒ s = 25/20 = 1.25 m
The height attained by stone, s = 1.25 m
We know that: v = u + at
⇒ 0 = 5 + (–10) × t
⇒ 0 = 5 − 10t
⇒ t = 5/10 = 0.5 s
Thus, the stone will attain a height of 1.25 m and the time taken to attain the height is 0.5 s.
Page No. 85
Q1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Ans: Here, the diameter of the circular track = 200 m.
The radius of the circular track, r = 100 m.
Let the athlete start moving from A, which is treated as a reference point.
Given:
Diameter of the circular track = 200 m
Radius, r = 100 m
Time for one round = 40 s
Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s
Circumference of the track = 2πr = 2 × (22/7) × 100 = 4400/7 ≈ 628.57 m
Speed = Circumference / Time = 628.57 / 40 ≈ 15.71 m/s
Distance covered in 140 s = 15.71 × 140 ≈ 2200 m
Number of rounds = Distance / Circumference = 2200 / 628.57 ≈ 3.5
After 3 complete rounds, the athlete is back at the starting point (e.g., point A). After 0.5 round more, the athlete is at the opposite point on the circular track (e.g., point B, 180° from A). Since the track has a diameter of 200 m, the displacement is the straight-line distance from A to B, which is the diameter:
Displacement = 200 m
Distance covered = 2200 m, Displacement = 200 m
Q2.Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in Jogging (a) from A to B and (b) from A to C?
Ans: Let Joseph jog from A to B and back to C as shown.
Given :
Distance from A to B = 300 m
Time from A to B = 2 minutes 50 seconds = 2×60+50=1702×60+50=170s
Distance from B to C = 100 m
Time from B to C = 1 minute = 6060s
Total distance from A to C = 300 m + 100 m = 400 m
Total time from A to C = 170 s + 60 s = 230 s
Displacement from A to B = 300 m (in the direction of motion)
Displacement from A to C = 300 m – 100 m = 200 m (since he jogs back 100 m)
Solution:
(a) From A to B:
(b) From A to C:
Q3. Abdul while driving to school computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 40 km h–1. What is the average speed for Abdul’s trip? Ans: Let the distance between Abdul’s home and school be d km.Speed while going to school = 20 km/hSpeed while returning home = 40 km/hTotal distance covered = d + d = 2d kmTime taken while going = d ÷ 20 hoursTime taken while returning = d ÷ 40 hoursTotal time taken = d ÷ 20 + d ÷ 40= (2d + d) ÷ 40= 3d ÷ 40 hoursNow,Average speed = Total distance ÷ Total time= 2d ÷ (3d ÷ 40)= (2d × 40) ÷ 3d= 80 ÷ 3 km/hFinal Answer:Average speed = 80⁄3 km/h or approximately 26.67 km/hQ4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms–2 for 8.0 s. How far does the boat travel during this time?Ans: Given, initial velocity of the boat = 0 m/s, Acceleration = 3 m/s² Time period = 8 seconds From the second equation of motion, s =ut+12at2 Thus, the total distance travelled by the boat in 8 seconds = 0 + 1/2 (3) (8)2 = 96 metersTherefore, the motorboat covers a distance of 96 meters in 8 seconds. Q5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? Ans:The total displacement of each car can be determined by calculating the area under the speed-time graph.For the first car, the displacement is given by the area of triangle AOB:=12×OB×OADisplacement of the first car= 1/2 ×OB×OAHere, OB=5 seconds and OA=52 km/h, which converts to 14.44m/s. Therefore, the area of triangle AOB is:12×(5)×(14.44)=36 metersFor the second car, the displacement is represented by the area of triangle COD:=12×OD×OCHere, OD=10 seconds and OC=3km/h, which converts to 0.83 m/s. Thus, the area of triangle COD is12×(10)×(0.83)=4.15 metersIn conclusion, the first car is displaced by 36 meters, while the second car is displaced by 4.15 meters. Therefore, the first car, travelling at 52 km/h, moved farther after applying the brakes.Q6. Figure below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions :(a) Which of the three is travelling the fastest?(b) Are all three ever at the same point on the road?(c) How far has C travelled when B passes A?(d) How far has B travelled by the time it passes C?Ans: (a) since the slope of line B is the greatest, B is travelling at the fastest speed.(b) since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.(c) since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.Since the initial point of the object, C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 kmWhen A passes B, the distance between the origin and C is 8kmTherefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km(d) the distance that object B has covered at the point where it passes C is equal to 9 graph units.Therefore, the total distance travelled by B when it crosses C = 9*(4/7) = 5.14 kmPage No. 86Q7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms–2, with what velocity will it strike the ground? After what time will it strike the ground?Ans: Initial velocity of the ball, u = 0 (as it is dropped)Height, h = 20 m, Acceleration, a = 10 ms-2,s = 20m a = 10m/s² t = ? s = ut + 1/2at² 20 = 0 × t + 1/2 × 10 × t² 20 = 0 + 5t² t² = 4 t = 2s v = u + at = 0 + 10 × 2= 20m/s OR v² = u² + 2asv² = 0 + 2 × 10 × 20 v² = 400 v = 20m/s The ball will strike the ground after 2 sec with the velocity of 20m/s.v = u + at20 = 0 + 10t∴ t = 20/10or time, t = 2 sTherefore, the ball reaches the ground after 2 seconds.Q8. The speed-time graph for a car is shown in figure below.(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.(b) Which part of the graph represents uniform motion of the car?Ans: (a)In the velocity-time graph,Distance = Area of the v-t graphHere we approximate the area using the area of the triangle,Distance travelled = Area of triangle AOBThe shaded area, which is equal to 1/2 × 4 × 6 = 12 m represents the distance travelled by the car in the first 4 s.(b)The part of the graph in red colour between times 6 s to 10 s represents the uniform motion of the car.Q9. State which of the following situations are possible and give an example for each of these.(a) An object with a constant acceleration but with zero velocity.(b) An object moving in a certain direction with acceleration in the perpendicular direction.(c) an object moving with acceleration but with uniform speed.Ans: (a) When an object is thrown upwards, it comes to a momentary rest at the highest point. Thus velocity is zero, but the acceleration due to the gravitational pull of the earth still acts upon it.(b) In a uniform circular motion, the speed remains constant, but there is varying velocity as it changes its direction, so there always be acceleration which is given by centripetal force.(c) When an object is thrown in the forward direction, then during its motion in the horizontal direction, the acceleration due to the gravity of the earth acts in the vertically downward direction.Q10. An artificial satellite is moving in a circular orbit with a radius of 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.Ans: Satellite completes one round in 24 hoursThe radius of the orbit r=42250 kmThe circumference C of the orbit is given by:C=2πrSubstituting the value of r:C= 2×3.1416×42250 km C≈ 265,571.6 km Speed v = C/TV= 265,571.6 km/24 hoursV ≈ 11, 065.48 km/h
Q1. What is a tissue? Ans: A group of cells that are similar in structure and/or work together to achieve a particular function forms a tissue.
Q2. What is the utility of tissues in multi-cellular organisms? Ans:
Tissues are made up of a group of cells carrying a specialised function. Each specialised function is taken up by a different tissue. Since these cells of a tissue carry out only a particular function, they do it very efficiently.
The use of tissues in multicellular organisms is to provide structural and mechanical strength.
Example: In human beings, muscle cells contract or relax to cause movement, nerve cells carry messages, and blood flows to transport gases, food, hormones, waste materials and so on. Likewise, in plants, vascular tissues (xylem, phloem) conduct water and food from one part of the plant to other parts.
So, multicellular organisms show a division of labour through tissues.
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Q1. Name types of simple tissues. Ans: Simple permanent tissues are of three types: (i) Parenchyma (ii) Collenchyma (iii) Sclerenchyma Parenchyma tissue is of further two types: (i) Aerenchyma (ii) Chlorenchyma
Q2. Where is apical meristem found? Ans: Apical meristem is present at the growing tips of stems and roots.
Q3. Which tissue makes up the husk of coconut? Ans: Sclerenchyma tissue makes up the husk of coconut. These tissues cause the plant to become stiff and hard. The cells of this tissue are dead and their cell walls are thickened because of the presence of lignin.
Q4. What are the constituents of phloem? Ans: The constituents of phloem are:
Sieve tubes
Companion cells
Phloem parenchyma
Phloem fibres
Page No. 69
Q1. Name the tissue responsible for movement in our body. Ans: Two tissues jointly are responsible for the movement of our body, namely:
Muscular tissue
Nervous tissue
Q2. What does a neuron look like? Ans: A neuron consists of a cell body with a nucleus and cytoplasm, from which long thin hair-like parts called dendrites arise. Each neuron has a single long part called the axon.
Q3. Give three features of cardiac muscles. Ans: Three features of cardiac muscles are: (i) Cardiac muscles are involuntary. (ii) Cardiac muscle cells are cylindrical, branched and uninucleate. (iii) Cardiac muscles show rhythmic contraction and relaxation.
Q4. What are the functions of areolar tissue? Ans: Areolar tissue acts as a supportive and packing tissue between organs lying in the body cavity, and also helps in the repair of tissues.
Page No. 70
Q1. Define the term “tissue”. Ans: A group of cells that are similar in structure and work together to achieve a particular function is called tissue.
Q2. How many types of elements together make up the xylem tissue? Name them. Ans: Xylem is composed of the following elements:
Tracheids
Vessels
Xylem parenchyma
Xylem fibres
Q3. How are simple tissues different from complex tissues in plants? Ans:
Q4. Differentiate between parenchyma, collenchyma and sclerenchyma, on the basis of their cell wall. Ans: The differences between the cell walls of parenchyma, collenchyma and sclerenchyma are:
Q5. What are the functions of the stomata? Ans: The functions of stomata are: (i) Stomata allow gaseous exchange between the plant and the atmosphere. (ii) These are sites of transpiration in plants.
Q6. Diagrammatically show the difference between the three types of muscle fibres. Ans: The three types of muscle fibres are: (i) Striated muscles(ii) Smooth muscles (unstriated muscle fibre)(iii) Cardiac muscles
Q7. What is the specific function of the cardiac muscle? Ans: The cardiac muscles are branched and cylindrical. They are uninucleated and are involuntary in nature. Throughout one’s lifetime, the cardiac muscles bring about the rhythmic contraction and relaxation of the heart.
Q8. Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body. Ans:
Q9. Draw a labelled diagram of a neuron. Ans: Q10. Name the following: (a) Tissue that forms the inner lining of our mouth. (b) Tissue that connects muscle to bone in humans. (c) Tissue that transports food in plants. (d) Tissue that stores fat in our body. (e) Connective tissue with a fluid matrix. (f) Tissue present in the brain. Ans: (a) Simple squamous epithelium (b) Tendon (c) Phloem (d) Adipose tissue (e) Blood (f) Nervous tissue
Q11. Identify the type of tissue in the following: Skin, Bark of Tree, Bone, Lining of Kidney Tubule, Vascular Bundle. Ans: Skin: Epithelial tissue (Squamous epithelium) Bark of Tree: Cork (protective tissue) Bone: Skeletal tissue (connective tissue) Lining of Kidney Tubules: Cuboidal epithelial tissue Vascular Bundle: Complex permanent tissue—xylem and phloem
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Q12. Name the regions in which parenchyma tissue is present. Ans: Parenchyma tissues are found in:
The pith of stems and roots.
When parenchyma contains chlorophyll it is called as chlorenchyma, it is found in green leaves.
Parenchyma found in aquatic plants have large air cavities that enable them to float and are hence called aerenchyma.
Q13. What is the role of the epidermis in plants? Ans: Role of the epidermis in plants: (i) It acts as a protective tissue, covering the plant body. (ii) It protects the plant from excessive heat or cold and from the attack of parasitic fungi and bacteria. (iii) It allows the exchange of gases and transpiration through stomata. (iv) The cuticle of the epidermis checks the excessive evaporation of water.
Q14. How does the cork act as a protective tissue? Ans: The cork cells are dead and do not have any intercellular spaces. The cell walls of the cork cells are coated with suberin (a waxy substance). Suberin makes these cells impermeable to water and gases, or in simpler words, it makes them waterproof and blocks gases from passing through. Thus, it protects underlying tissues from desiccation (loss of water from the plant body), infection and physical damage.
American War of Independence
Windmills harnessing the Kinetic energy
). Hence, it will sink in water.
The bus stops suddenly, Passenger jerks forward
