Q.2. निम्नलिखितस्य श्लोकस्य भावार्थं हिन्दीभाषया आङ्ग्लभाषया वा लिखत- यो मां पुत्रप्रीत्या पोषयति स्वामिनो गृहे तस्य। रक्षानियोगकरणान्न मया भ्रष्टव्यमीषदपि।। उत्तरम्- सरलार्थ – ये पक्षी मनुष्य के पास नहीं जाते हैं। (अतः) वह बालक उदास हो गया। अतः मनुष्य के उचित मनोरंजन करने वाले को खोजता हूँ। इस प्रकार घूमकर दौड़ते हुए किसी कुत्ते को देखा। प्रसन्न होकर बालक ने उसे इस प्रकार कहा-अरे मनुष्यों के मित्र। इस गर्मी के समय में क्यों घूम रहे हो? छाया के कारण शीतल इस वृक्ष की जड़ के पास आश्रय ले लो। मैं भी तुम्हें खेल का उचित साथी मानता हूँ। कुत्ता कहने लगा-जो (व्यक्ति) मुझे पुत्र जैसे स्नेह से पुष्ट करता है, उस स्वामी के घर में रक्षा कार्य से जरा भी मुझे हटना नहीं चाहिए।
Q.3. स्थूलपदान्यधिकृत्य प्रश्ननिर्माणं कुरुत- (क) स्वादूनि भक्ष्यकवलानि ते दास्यामि। (ख) चटकः स्वकर्मणि व्यग्रः आसीत्। (ग) कुक्कुरः मानुषाणां मित्रम् अस्ति। (घ) स महती वैदुषर्षी लब्धवान् (ङ) रक्षानियोगकरणात् मया न भ्रष्टव्यम् इति। उत्तरम्-: (क) कानि भक्ष्यकवलानि ते दास्यामि? (ख) चटकः कस्मिन् व्यग्रः आसीत् ? (ग) कुक्कुरः केषां मित्रम् अस्ति? (घ) स कां लब्धवान् ? (ङ) कस्मात् मया न भ्रष्टव्यम् इति?
Q.3. अधोलिखितयोः श्लोकद्वयोः आशयं हिन्दीभाषया आङ्ग्लभाषया वा लिखत- (क) आरम्भगुर्वी क्षयिणी क्रमेण लघ्वी पुरा वृद्धिमती च पश्चात् । दिनस्य पूर्वार्द्धपराद्धभिन्ना छायेव मैत्री खलसज्जनानाम् ॥ (ख) प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः। तस्मात्तदेव वक्तव्यं वचने का दरिद्रता ॥ उत्तरम्- (क) सरलार्थ- दुष्ट और सज्जन लोगों की मित्रता क्रमशः दिन के पूर्वार्द्ध और परार्द्ध की छाया की तरह प्रारम्भ – में विशाल तथा शनैः शनैः घटने वाली होती है और आरम्भ में लघु तथा पश्चात् वृद्धि वाली होती है। भाव यह है कि जिस प्रकार दिन के पूर्वार्द्ध की छाया प्रारम्भ में विशाल तथा बाद मे घटती जाती है। उसी प्रकार दुष्ट लोगों की मित्रता प्रारम्भ में विशाल तथा अन्त में छोटी होती जाती है। इसके विपरीत सायं की छाया प्रारम्भ में लघु होती है तथा पश्चात् बढ़ती जाती है। इसी प्रकार सज्जन लोगों की मित्रता प्रारम्भ में लघु और बाद में प्रगाढ़ होती जाती है। उत्तरम्- (ख) सरलार्थ- सभी प्राणी मधुर वचनों के प्रदान से प्रसन्न होते हैं। अतः वैसा ही बोलना चाहिए। (मधुर) बोलने में कैसी कंजूसी?
प्रश्न 5: प्रथमश्लोकस्य आशयं हिन्दी भाषया आङ्गल भाषया वा लिखत – उत्तर: प्रथमश्लोकस्य आशय – हे सरस्वती! नवीन वीणा बजाओ, सुन्दर नीतियों से युक्त मधुर गीत गाओ।
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.
