AP Board 7th Class Science Solutions 12th Lesson Soil and Water

SCERT AP 7th Class Science Study Material Pdf 11th Lesson Soil and Water Textbook Questions and Answers.

AP State Syllabus 7th Class Science 12th Lesson Questions and Answers Soil and Water

7th Class Science 12th Lesson Soil and Water Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Wastewater released from houses is called _______.
2. The Science dealing with the formation of soil is called _______.
3. In modern water purifiers _______ are used instead of chlorine to kill the germs.
Answer:
1. Sewage
2. Pedology
3. Ultraviolet rays

II. Choose the correct answer.

1. In addition to the soil particles, the soil also has
a) air and water
b) water and plants
c) minerals, organic matter, air and water
d) water, air and plants
Answer:
c) minerals, organic matter, air and water

2. The water holding capacity is the highest in
a) sandy soil
b) clayey soil
c) loamy soil
d) mixture of sand and loam
Answer:
b) clayey soil

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

3. Which among the following is not responsible for water shortage?
a) Industrial growth
b) Population growth
c) Heavy rainfall
d) Mismanagement of water resources
Answer:
c) Heavy rainfall

III. Matching.

A) Weathering1. more water retaining capacity
B) Clay soil2. washing off the top soil
C) Soil erosion3. soil formation
D) Afforestation4. 1%
E) Fresh water5. growing trees
6. 99%

Answer:

A) Weathering3. soil formation
B) Clay soil1. more water retaining capacity
C) Soil erosion2. washing off the top soil
D) Afforestation5. growing trees
E) Fresh water4. 1%

IV. Answer the following questions

Question 1.
Explain briefly the terms.
i) Weathering ii) Aquifer iii) Percolation iv) Sewage
Answer:
i) Weathering :

  1. In nature due to the action of various natural agents such as wind, water, sun and climate the bigger rocks (parent rock) gradually breakdown and give small particles. These form the soil. This process is known as Weathering.
  2. It takes approximately 500 -1000 years for the formation of 1 inch of soil.
  3. Weathering helps in the formation of soil.
  4. The process of formation of soil from the parent rock by the process of weathering is called ‘Pedogenesis’.

ii) Aquifer :

  1. Ground water is stored between layers of hard rock below the water table. This is called aquifer.
  2. The wells, tube wells and hand pumps get water present in the aquifer.

iii) Percolation :

  1. The absorption and downward movement of water through the soil layer is called percolation.
  2. The rate of percolation various from soil to soil.
  3. Sandy soils are well aerated and they have high percolation rate.

iv) Sewage :

  1. All the waste water released by homes, industries, hospitals, offices and other users are collectively called sewage.
  2. Sewage is a liquid waste.
  3. Sewage contain water with dissolved and suspended impurities, disease causing bacteria and other microbes. These impurities are called contaminants.
  4. Discharge of untreated sewage into canals causes harm to aquatic life.

Question 2.
List the differences between clayey soil, loamy soil and sandy soil. How is clayey soil useful for crops?
Answer:
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 1

  1. Clayey soils are less aerated and their water holding capacity is higher.
  2. It has finer particles less than 0.002 mm. Hence, they retain the water for a longer period.
  3. Clayey soils are very suitable to grow crops like paddy. Sugarcane and cotton. Black soil with highest clay are suitable to grow crops like redgram. chillies etc.

Question 3.
Harshith observed that majority of the houses in his locality get water through bore well but no house has taken measures for rain water harvesting. Guess the long term effect on the water table.
Answer:

  1. Over use of ground water can cause wells to dry up.
  2. This leads to expensive and ultimately fertile attempts to keep up with the dropping water table by drilling deeper and deeper wells.
  3. People will have to travel for miles to get water for their daily use.
  4. People will migrate to longer distances in search of drinking water and villages would become barren.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 4.
If you get a chance to interview a soil scientist, what questions will you ask related to soil, its testing and conservation.
Answer:
I shall ask the following questions.

  1. How can you test the soil for its acidity?
  2. Is it possible to change the basic nature of the soil?
  3. How can the water holding capacity of the soil be increased at a low cost?
  4. Suggest methods to improve the fertility of the soil.
  5. What can be done with the black cotton soil when a building is to be constructed?
  6. As black cotton soil yields, and the building cracks, suggest preventive measures for it.

Question 5.
Your friends conducted an experiment in the field regarding the rate of percolation. They observed that it took 40 min for 200 ml of water to percolate through the soil. Calculate the rate of percolation. Also write the procedure to conduct the test.
Answer:
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 2
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 3

Procedure :

  1. Take a soil sample.
  2. Take a plastic funnel and place a filter paper in it as shown in the figure.
  3. Weigh 50 gms of dry soil and pour it into the funnel.
  4. Take 200 ml of water in a measuring cylinder.
  5. Then pour the water on the soil drop by drop.
  6. Pour the water all over the surface of the soil until it starts dripping out of the funnel. Calculate the time taken for the last water drop passes down the soil.

Question 6.
We call earth as “Bhoomatha -The Mother Earth”. How do you express your appreciation and gratitude to her?
Answer:

  1. Soil is one of the most important natural resources. It supports the existence of living organisms.
  2. We use soil for different purpose in our daily life. Almost all living things in our surroundings directly or indirectly depend on soil.
  3. The soil supports all plants, animals and micro organisms.
  4. We grow our food components in this soil.
  5. Soil is the birth place of all minerals, everything we get from this soil. Hence, we consider the earth as “Bhoomatha – The Mother Earth”.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 7.
Poorvika wants to maintain a garden,- what suggestions can you give her to minimize the use of water and also to improve the soil quality?
Answer:

  1. By using sprinklers and drip irrigation methods we can reduce the wastage of water in the garden.
  2. Construction of soaking pits in the garden enhances the percolation of raia water into the soil. It improves the ground water table.
  3. Rotting vegetation and animal remains make up the humus rich in fertility.
  4. They contain nutrients, natural manure, which give good support to the growth of ‘ plants.
  5. I also suggest Poorvika to grow big trees around the garden to stop winds.
  6. It is better to grow grass and other plants in the vacant portions of the garden. These grass plant roots hold the solid particles and prevent soil erosion during heavy rains.

7th Class Science 12th Lesson Soil and Water InText Questions and Answers

7th Class Science Textbook Page No. 84

Question 1.
Where do animals and humans live?
Answer:
They live on the Earth.

Question 2.
What would happen if there is no soil on the Earth?
Answer:
Life on the earth would become impossible.

Question 3.
Why is woman in the picture using hand pump?
Answer:
For getting drinking water.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 4.
Why is soil an important resource?
Answer:
It is the important natural resources necessary for life on the earth. Human beings, plants and micro organisms depend on soil for their basic needs.

Question 5.
Where do plants get nutrients from?
Answer:
From soil.

Question 6.
Did you ever see earthworms and snails coming out of soil during rainy season?
Answer:
Yes, they live in the soil by making burrows.

7th Class Science Textbook Page No. 85

Question 7.
Is soil useful to us in any other way?
Answer:
Yes, soil is also needed.

  1. For agriculture
  2. For constructing buildings
  3. For making toys and idols
  4. For making utensils and pottery
  5. For making cosmetics (multani soil)

Question 8.
Did you observe stones of different sizes at the bank of a stream?
Answer:
Yes. I observed many stones with different sizes and shapes.

Question 9.
What happens when rocks get rubbed and scroll due to the flow of water?
Answer:
Weathering of rocks takes place.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 10.
You might have observed white coloured crinkles on the rocks at the bank of river. How do they appear?
Answer:
They appear like shining jewels. They are formed by the disintegration of rock due to the abarsion of water stream.

Question 11.
Have you observed the top soil in your surroundings? What is it made of?
Answer:
Yes. It is made of dead and decayed organic matter called Humus.

7th Class Science Textbook Page No. 86

Question 12.
What do you observe floating on the top?
Answer:
Humus.

Question 13.
What do you find at the bottom of the beaker?
Answer:
Inorganic matter of the soil.

Question 14.
Do you find any insects and plant parts in the beaker?
Answer:
Yes, they are floating. It is the organic matter of the soil.

Question 15.
What do you infer from your observation?
Answer:
We can infer that soil consists of water, air, organic matter, inorganic matter and organisms.

Question 16.
How many layers did you observe in the pit?
Answer:
Two layers can be observed in the pit.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 17.
Do all the layers have same colour, texture and depth?
Answer:
No, all the layers have different colour, texture and depth.

7th Class Science Textbook Page No. 87

Question 18.
How can you identify which soil is suitable for this purpose?
Answer:
Clayey soil is suitable for making Ganesh idol.

7th Class Science Textbook Page No. 88

Question 19.
Do all types of soils have same properties?
Answer:
No, they have different properties.

7th Class Science Textbook Page No. 89

Question 20.
What type of soils are generally seen in our state?
Answer:
Sandy soils, black soils, red soils, loamy soils are seen in our state.

Question 21.
Can we grow a crop in all types of soil?
Answer:
No, it is not possible.

Question 22.
What type of soil is required for growing paddy?
Answer:
Paddy is grown in clayey soil, which retains water for a longer period.

Question 23.
Is there any relationship between the crops and soil?
Answer:
Yes, all crops are not grown in all soils. Crops are soil specific. For example cashew is grown in sandy soils and cotton is grown in black soils.

7th Class Science Textbook Page No. 90

Question 24.
Why is soil testing done?
Answer:
It helps a farmer to make a soil suitable for growing crops.

Question 25.
How do the farmers benefit by soil testing?
Answer:

  1. Soil testing helps the farmer to know about the current health of the farm’s soil and how to improve it.
  2. It helps farmers to prevent the degradation of soil quality.
  3. It helps to minimize usage of fertilizers.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 26.
Did you observe that after heavy rain, the top soil get washed off?
Answer:
Yes. I observed it during rainy season.

Question 27.
What will happen if the top soil is washed off?
Answer:
The soil will loose its quality and fertility.

Question 28.
How can we prevent it?
Answer:

  1. By afforestation. Controlling the deforestation activities.
  2. Constructing checkdams and bunds
  3. Preventing over grazing of land by animals.
  4. By planting trees as shelter belt to stop winds. .
  5. By planned agricultural practices we can prevent soil erosion successfully.

7th Class Science Textbook Page No. 91

Question 29.
What are the other causes of soil erosion?
Answer:

  1. Unplanned and excessive usage of agro chemicals
  2. Over grazing of land
  3. Deforestation
  4. Digging the earth for construction and mining

7th Class Science Textbook Page No. 92

Question 30.
Is water from seas and oceans useful for drinking and agriculture?
Answer:
No, the sea water is not useful for drinking and agriculture. But by desalinating the sea water we can use them for consumption.

Question 31.
Where is fresh water available from?
Answer:
From rivers, lakes, streams and ravines.

Question 32.
How much percentage of fresh water is available?
Answer:
only 1%.

Question 33.
How does the water reach the ground?
Answer:
The water reach the ground by infiltration.

Question 34.
Is the water table level same at all places?
Answer:
No, it is always vary in different places.

7th Class Science Textbook Page No. 93

Question 35.
Why do wells dry up?
Answer:
Decrease in ground water table leads to dry up of wells.

Question 36.
What will happen if the ground water table go down?
Answer:
All the wells and bore wells will dry up.

7th Class Science Textbook Page No. 94

Question 37.
Why is it essential to manage the water?
Answer:
To resolve the prevailing water scarcity in the society, proper water management of water resources is very essential.

Think & Respond

Question 1.
You may come across some people breaking large rocks into smaller ones to make road. Is it also the weathering of rocks? Why?
Answer:

  1. In natural weathering process, the parent rock is gradually breakdown into small particles by the action of air, water, sun heat, and climate. It takes 500 to 1000 years times. This helps in formation of soil.
  2. The break do\yn of rocks for the purpose of constructing roads and buildings also comes under weathering.

Here, the big rocks are blown down by using dynamites. In very short time, the bigger rocks become small gravel. It is artificial weathering or man made weathering. But this will not help in the formation of soil. This leads to the rapid depletion of natural resources like hills and mounds. The natural weathering factors like wind, water, sun have no role in this artificial weathering.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Question 2.
There is huge amount of water on earth, then why is water called a precious resource?
Answer:

  1. About 97% of water is present in seas and oceans which is not useful for drinking and agriculture.
  2. 2% of water is present in the form of rain, snow, sleet and precipitated water which is not available for consumption.
  3. Hence only 1% of fresh water is available in the form of surface and ground water sources. This 1% of water is useful for drinking, domestic usage and agriculture.
  4. Hence, we have to use this 1% of fresh water wisely and judiciously. This water is basis of life and human civilization. So, water is life. Save water save lives.
  5. So, we considered water as a precious resource.

Activities and Projects

Question 1.
Under the guidance of your teacher collect the wet waste from Mid Day Meal preparation and make Vermi compost for your school garden.
Answer:
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 4

  1. With the help of my science teacher, we constructed 10 × 1 × 1/2 meters Vermi Compost beds in sheds which protect these beds from direct sunlight and rain.
  2. We collected coconut, banana and sugarcane leaves. We made them into 3- 4 inches layer. This inner layer was wet with water.
  3. We collected the wet waste from Mid Day Meal to fill the bed. We were careful to avoid glass, polythlene, rubber and metal objects in the bed.
  4. After two weeks of making bed we kept thousand earth worms per square meter and covered the bed with Gunny bags to maintain 30 to 40% of moisture.
  5. After 60 days we collected the first manure. Second time we got the manure with in 40 – 50 days.
  6. By using this Vermi Compost we can reduce the investments on chemical fertilizers and other pesticides. The quality of the soil and agricultural produce increased.

Question 2.
Prepare a model of representing various horizons in the soil profile using a disposed plastic bottle, stones and soil.
Answer:
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 5

  1. Take soil from your surroundings, clean d: posable plastic bottle and home made paper funnel.
  2. Pour the soil into the funnel putting it on the plastic bottle.
  3. Now pour some water into the bottle upto its neck. Close the end of the bottle and shake it well. The heavier particles like gravel and sand settle to the bottom of the bottle, wait for a day to get all the soil particles to settle down in the bottle.
  4. After one day, you may observe that heavier particles like gravel settle at the bottom of the bottle. Above that you may find sand particles. Just above the sand particles, you may find very finer particles on the top of the bottle you will find humus floating on the surface of the water.
  5. From the top, first you find humus, the plant and animal organic matter which helps the growth of plants.
  6. Just below the hurhus, you find clay. Clay traps the water in the soil for the roots to absorb them.
  7. Below the clay, sand and gravel, they help to drain the water quickly so that the roots don’t rot.
  8. Basing on the percentage of each individual component loam soils may be classified into sandy loam, silty loam and clay loam.
  9. Here, the soil taken for this project is sandy loam. Like this way, we can determine the type of soils in our surroundings.

Activities

Activity – 1

Question 1.
How do you observe the plants and small living organisms in a given soil? Write your observations in a table.
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 6
Answer:

  1. I visited my school garden and the kitchen garden yesterday to observe the plants and small living things in those places.
  2. I marked an area of 30 cm × 30 cm. 1 dug the soil to a depth of 4 to 6 cm in the marked area.
  3. I carefully sorted out the soil and observed the presence of plants and small living organisms with the help of a hand lens.

I noted down the things 1 found in the given table.

Inference :

  1. We can infer that soil consists various plants and dried plant parts etc.
  2. It is a good habitat for many small organisms.

Activity – 2

Question 2.
How do you observe the different components of the soil?
Answer:
Aim : To observe the different components of the soil.

Materials required : Soil, beaker, water, polythene bag.

Procedure :
AP Board 7th Class Science Solutions 12th Lesson Soil and Water 8

  1. Collect a handful of soil sample in a transparent polythene bag and close it tightly.
  2. Leave it under the sun for 2 hours we can observe some droplets of water on the inner side of the bag.
  3. Fill a beaker with this handful of soil, pour water into it slowly and carefully. Fill the beaker with water and stir well the soil and water.
  4. Leave it for undisturbed for some time.

Observations :
Organic substances, dry rotten leaves and roots float on water. Dead and decayed organic matter that mixes with soil is called humus.
My observations and inferences are shown in the table.

ObservationInference
Droplets of water in the bagPresence of water in soil
Bubbles from soil when water is pouredPresence of air in soil
Floating of dry plant partsPresence of organic matter in soil
Particles at the bottomPresence of inorganic matter in soil
Insects and plant partsPresence of organisms in soil

Inference :
Soil consists of water, air, organic matter, inorganic matter and organisms.

Activity – 3

Question 3.
How do you identify the type of soil in your science lab?
Answer:

  1. We can identify the different types of soils in the line diagram given below.
  2. We can determine the soil on the basis of following chart.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water 9

Activity – 4

Question 4.
How do you know about soil erosion?
Answer:
Aim : To know about soil erosion.
AP Board 7th Class Science Solutions Chapter 12 నేల మరియు నీరు 6

Procedure :

  1. Cut three disposable bottles sidewise and fill with soil as shown in the figure.
  2. Put some sprouts of green gram or any other seeds in one bottle and water regularly.
  3. Cover the second bottle with dry leaves and leave the third bottle without sowing anything.
  4. After a week, the sprouts in the first bottle will grow more.
  5. Now, make arrangements to collect the drained water from each bottle by arranging small vessels.
  6. Now blow air and pour water slowly in each bottle.

Observation :
In the first bottle containing sprouts, less soil is blown out by air and washed off by water.

Inference :
We can infer that the top soil is washed off by water and blown out by air. Plants help to prevent soil erosion.

AP Board 7th Class Science Solutions 12th Lesson Soil and Water

Activity – 5

Question 5.
What are the differences in the water table in your locality due to increase in the construction and industries nearby?
Answer:

  1. The water levels in the wells are rapidly decreased.
  2. Many borewells are not working due to depletion in the water table.
  3. People are drilling deeper and deeper borewells to keep up with the dropping water table. It is an expensive task. Hence, they are facing financial troubles.
  4. People are facing drinking water shortage in the village.
  5. People have to walk for miles to get water for their usage.
  6. Many people migrated to different places due to scarcity of drinking water in the village.
  7. Population explosion, industrial growth, enhanced agriculture activities, deforestation, decrease in the effective area for seepage and drop in rainfall are other main reasons for the rapid decrease of water table on the earth.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

SCERT AP 7th Class Science Study Material Pdf 10th Lesson Changes Around Us Textbook Questions and Answers.

AP State Syllabus 7th Class Science 10th Lesson Questions and Answers Changes Around Us

7th Class Science 10th Lesson Changes Around Us Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Changes in which new substances are formed are called ______ changes.
2. Magnesium + Oxygen → ______
3. Milk converted to curd is ______ change.
4. Rusting of iron is protected by ______ process.
Answer:
1. Chemical
2. Magnesium Oxide
3. Chemical
4. Galvanisation

II. Choose the correct answer.

1. When a woolen yarn is knitted to get a sweater, the change can be described as
a) Physical change
b) Chemical change
c) Endothermic reaction
d) Exothermic reaction
Answer:
a) Physical change

2. The chemical chancre among the following is ______
a) Water to clouds
b) GroWth of a tree
c) Cow dung to bio gas
d) Ice to water
Answer:
b) GroWth of a tree

3. Which is an example of a periodic change?
a) Earth quake
b) Formation of rainbow
c) Occurrence of tides in sea
d) Showering of rain
Answer:
c) Occurrence of tides in sea

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

4. Photosynthesis by green plant is a
a) Physical change
b) Chemical change
c) Reversible change
d) None of the above
Answer:
b) Chemical change

III. Matching.

Group – AGroup – B
A) Growing hair1. Chemical change
B) Breaking Mirror2. Acetic Acid
C) Galvanisation3. Slow change
D) Vinegar4. Physical change
E) Atmospheric pollution5. Depositing zinc on iron metal
6. Fast change

Answer:

Group – AGroup – B
A) Growing hair3. Slow change
B) Breaking Mirror4. Physical change
C) Galvanisation5. Depositing zinc on iron metal
D) Vinegar2. Acetic Acid
E) Atmospheric pollution1. Chemical change

IV. Answer the following questions.

Question 1.
Distinguish physical and chemical changes.
Answer:

Physical ChangesChemical Changes
1) No new substances are formed.1) During this new substances are formed.
2) The chemical properties of a sub-stance do not change.2) The chemical properties of a substance changes.
3) It is temporary and reversible in nature.3) It is permanent and irreversible change.
4) Change in the properties such as colour, shape and size of a sub-stance occurs.4) A colour change may take place and sound may be produced. Heat or light may be ab¬sorbed or released.

Question 2.
When a candle is burnt, what type of changes takes place? Give another example of the similar process.
Answer:

  1. When a candle is burnt both physical and chemical changes take place.
  2. Burning of candle giving heat and light and reducing its size is a chemical change.
  3. The wax melted during burning again condenses to solid wax is a physical change.
  4. Part of the candle burnt is a permanent change and it cannot be reversed.
  5. Due to the burning of the candle, carbon dioxide and water vapour are produced.

Examples:

  1. Burning of LPG in our kitchen.
  2. Liquid state of LPG in the cylinder is converted into gaseous state when comes out This is a physical change.
  3. When LPG burns in air heat is produced, this is a chemical change.

Question 3.
Define crystallization.
Answer:
The process of separating a soluble solid from the solution by heating or evaporating is called crystallization.

Question 4.
Guess the consequences of burning crackers during festivals and celebrations.
Answer:

  1. Firing crackers during festivals and celebrations increase the concentration of dust and pollutants in the air.
  2. The dust particles get settled on the surrounding surfaces which are packed with chemicals like copper, zinc, sodium and magnesium. They cause damage to paintings.
  3. The quality of ambient air will be decreased drastically.
  4. Firing of crackers causes lot of sound pollution.
  5. The oxides of sulphur pollute the air and causes lung diseases to human beings.
  6. The increase in levels of air pollutants cause acid rains, corrosion of objects and decrease in visibility.

Question 5.
Prove experimentally chemical change is a permanent change.
Answer:
AP Board 7th Class Science Solutions Chapter 10 Changes Around Us
Aim : To prove chemical change is a permanent change.

Materials required :
Magnesium ribbon, water, match box, bunsen flame.

Procedure:
1) Take a small piece of Magnesium ribbon.

2) Burn it on a flame we will find brilliant white dazzling light leaving a powdery substance behind.

3) This is because when magnesium burns in the presence of oxygen, it forms magnesium oxide in the form of powdered ash. This is a new substance.
Magnesium + Oxygen → Magnesium Oxide (new substance)
This is a permanent change.

4) Now, take a small quantity of Magnesium oxide and mix it with a small quantity of water and dissolve it.
Magnesium Oxide + Water → Magnesium Hydroxide (new substance)
This is also a permanent change.
Here also another new substance is formed.

5) By testing this new substance with blue and red litmus papers. Red litmus turns blue. Then Magnesium Hydroxide is a base.

Inference :
Changes that occur with the formation of new substance with different chemical Coposition or transformation of a substance into another substance with the evolution or absorption of heat; or light energy are termed as chemical changes. These are permanent changes.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 6.
Guess the reasons for increase of plastic pollution.
Answer:
Main reasons for increase of plastic pollution :

  1. Rising human population and their needs, people depend on plastic.
  2. The food industry which packs everything on plastic.
  3. Increase in usage of plastic bottles and container caps.
  4. Indiscriminate usage of plastic bags and carriers.
  5. Over usage of plastic straws and stirrers.
  6. Lack of proper management of plastic waste.
  7. Unawareness among the public about the hazardous affects of plastic wastes.
  8. People are not following the 3’Rs (Recycle – Reduce – Reuse) method to reduce plastic wastes.

Question 7.
How can you appreciate the role of periodic changes in nature?
Answer:

  1. Changes that are repeating at regular intervals of time are called as periodic changes.
  2. Formation of day and night, for every 12 hours the sun rise and the sun set are repeating.
  3. Seasons like summer, rainy, winter and spring occur every year at regular intervals.
  4. The full moon and new moon repeats every month.
  5. Heart beat of human beings takes place periodically.
  6. Low and high tides in sea are periodic.
  7. All these changes are periodically taking place in nature.
  8. They provide us all comforts to live on this earth. I appreciate these periodic changes in nature for providing a happy life on this earth.
  9. Without these changes in environment, life of us would become impossible.

Question 8.
Suggest some methods to prevent rusting of iron articles.
Answer:

  1. The problem of rusting of iron and other metal articles is common experience in almost every home.
  2. It spoils beautiful articles and makes them look ugly.
  3. The following are some of the ways to prevent the rusting of iron.
    a) Do not allow the iron articles to come in direct contact with Oxygen in the air, water or both.
    b) Apply a coat of paint or grease on an iron article.

Question 9.
Ravi prepared carbon-dioxide using baking soda and vinegar. Carbon-dioxide changed lime water into milky white. Represent this experiment in a diagram with labelling.
Answer:
AP Board 7th Class Science Solutions 10th Lesson Changes Around Us 1
Preparation of carbon dioxide

7th Class Science 10th Lesson Changes Around Us InText Questions and Answers

7th Class Science Textbook Page No. 54

Question 1.
What change do you observe after removing mehandi applied on your palms?
Answer:
The hands get red colour after removing mehandi.

Question 2.
Can we clear the printed letters on paper. Why?
Answer:
No, it is a permanent change.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 3.
Water dispenser gives hot, normal and cold water. Is this natural or man made?
Answer:
Man-made.

7th Class Science Textbook Page No. 55

Question 4.
What is change?
Answer:
An act or process through which something becomes different.

Question 5.
How do we know that something has changed?
Answer:
By observing the changes in shape, size, colour and chemical properties.

Question 6.
What are the possible reasons for that change?
Answer:
They may be because of physical or chemical changes.

Question 7.
Is there any involvement of human beings?
Answer:
No, these are natural changes.

7th Class Science Textbook Page No. 56

Question 8.
Is there any change in its shape?
Answer:
Yes, it’s size will increase.

Question 9.
How does it happen?
Answer:
The air blown into the balloon occupies the space and the balloon becomes bigger in size. ,

Question 10.
Is it by itself or by anybody else?
Answer:
It is a man made change.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 11.
So what type of change it is?
Answer:
This is a man made change.

Question 12.
How much time they require?
Answer:
They require longer periods. It is a slow process.

7th Class Science Textbook Page No. 57

Question 13.
Have you ever noticed the following changes in our daily life?
Answer:
Yes. It is also a reversible change.

7th Class Science Textbook Page No. 58

Question 14.
Do the formed substance differs from original substance?
Answer:
Yes. The formed new substance is completely different.

Question 15.
Can we get original substance by reversing the conditions?
Answer:
No.

7th Class Science Textbook Page No. 50

Question 16.
Have you seen large crystals of sugar (Missrf) or crystal salt?
Answer:
Yes.

Question 17.
Do you know how we get these crystals?
Answer:
Yes. We get them through crystallization.

Question 18.
Have you ever observed the formation of small sugar crystals on sweets like Jilebi and badushah, which are kept aside for a long period?
Answer:
Yes.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 19.
What is the reason for this?
Answer:
It is because of crystallization.

Question 20.
What changes do you notice at the end?
Answer:
We notice the formation of large size crystals of sugar at the bottom of the beaker.

Question 21.
Do you find any crystals in the solution?
Answer:
No, we will find the crystals at the bottom of the beaker.

Question 22.
What type of change it is?
Answer:
It is a physical change.

7th Class Science Textbook Page No. 62

Question 23.
In Which cases new substances are formed?
Answer:
In chemical changes, we get new substance.

Question 24.
Have you observed rusting of iron, curdling of milk?
Answer:
Yes, I observed many times.

Question 25.
Do we get the same substance after change? Is the change is temporary or permanent?
Answer:
No, it is a permanent change.

Question 26.
Does the ash formed look like Magnesium ribbon?
Answer:
No.

Question 27.
Do you think the magnesium ribbon and the ash have the same composition?
Answer:
No

Question 28.
What do you observe?
Answer:
We get Magnesium Hydroxide.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 29.
Do you observe any change in the state of the substance?
Answer:
It is in liquid state (from solid state)

Question 30.
Is it an acid or base?
Answer:
Base.

7th Class Science Textbook Page No. 63

Question 31.
Did you observe iron nails, iron gates, iron benches or pieces of iron left in the open ground for a long time?
Answer:
Yes.

Question 32.
Did you observe any change in colour?
Answer:
We can observe a brown colour layer on the surface of iron articles.

Question 33.
How can we protect the iron articles from rusting?
Answer:
Through a process called Galvanization.

7th Class Science Textbook Page No. 64

Question 34.
Are there any other ways by which rusting of iron can be prevented?
Answer:
When the iron articles are coated with metals like chromium or zinc, we can prevent them from rusting.

Question 35.
Have you observed the handles of bicycles, metal rims of bicycle and motor cycles?
Answer:
Yes.

Question 36.
Do these articles rust? If not why?
Answer:
No, because they are coated with zinc or chromium.

Question 37.
What changes do you notice?
Answer:
We can notice a brown layer on their outer surface.

Question 38.
Can you prevent the browning of cut vegetables and fruits?
Answer:
Yes.

Question 39.
Have you observed your mother keeping the cut potatoes or brinjals in cold water?
Answer:
Yes.

7th Class Science Textbook Page No. 65

Question 40.
In which fruit or vegetable do you notice change in colour?
Answer:
In Apple, Brinjal and Potato.

Question 41.
Why does this change occurs?
Answer:
Because some vegetables and fruits react with oxygen in the air when they cut. Due to this oxidation process, brown layer is formed on the surface of these fruits and vegetables.

Think & Respond

Question 1.
When food gets spoiled, it produces a foul smell. Shall we call this change as a chemical change?
Answer:
When food gets spoiled by micro organisms, it produces a foul smell. By the action of the enzymes released by these microbes, the components of the food undergo chemical changes. Due to this, the food emits foul smell. It is a permanent change and irreversible in nature.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Question 2.
You know that plants produce their food by a process called photosynthesis. Can we call photosynthesis a chemical change?
Answer:
Photosynthesis is a chemical change. Here, carbondioxide is reduced to glucose. In this process oxygen is evolved. It is a permanent and irreversible change. New substance are formed.

Question 3.
Do all the materials react with oxygen in the air?
Answer:
No.

Question 4.
Observe Gold and Silver. You wear them in the form of ornaments. Even if they get exposed to air for long time, they do not change their colour. Why?
Answer:
Metals like gold and silver, even if they are exposed to air for a long time, they do not change colour or get rusted.

It means that they are resistant to corrosion which is the reason why we use them in making ornaments.

Activities and Projects

Question 1.
Collect information on the process of artificial ripening of fruits in fruit market and discuss whether it is useful or harmful.
Answer:

  1. Calcium carbide more commonly known as “masala” is used for artificial ripening of fruits. It is very harmful to health.
  2. Calcium carbide is a carcinogenic agent.
  3. The most important precaution to avoid eating such artificially ripened fruits is to go in for fruits and vegetables which are not unseasonal.
  4. Always wash the vegetables and fruits properly before consuming.

Question 2.
When you burn a piece of wood different changes take place analyse the following. Predict possible changes and list them all.
a) Are there any physical changes among them?
b) How many forms of energy are released in the change?
c) What chemical changes do you notice?
d) Explain briefly why these occur.
Answer:
a) No physical changes is observed.
b) The energy is released in the form of heat sound and light energy.
c) When a piece of wood is burnt a new material is formed (ash). We also notice change in shape and size of new materials (powder). This type of change which leads to form new substance is called chemical changes.
d) Wood turns into carbondioxide and ash
Carbon (Wood) + Oxygen (Air) → CO2 + Ash

Activities

Activity – 1

Question 1.
What changes do you observe, When a balloon is blown?
Answer:
Change in the shape of balloon is done by blowing air into it. This is a man made change. When we left off the air from it, it returns to its normal shape. This is a physical change and reversible one.

Activity – 2

Question 2.
Observe the changes mentioned in the following table and complete it.

ChangeTime taken
shorter/longer duration
Type of change
Fast/ Slow
1. Digestion of food
2. Burning a small candle
3. Occurrence of lightening
4. Construction of Dam
5. Rusting of iron

Answer:

ChangeTime taken
shorter/longer duration
Type of change
Fast/ Slow
1. Digestion of foodLonger durationSlow
2. Burning a small candleShorterFast
3. Occurrence of lighteningShorterFast
4. Construction of DamLonger durationSlow
5. Rusting of ironLonger durationSlow

Activity – 3

Question 3.
What changes do you notice when few pieces of ice in a beaker are heated? What do you mean by a physical change? Before heating After heating After freezing
Answer:
AP Board 7th Class Science Solutions 10th Lesson Changes Around Us 2

  1. Take few pieces of ice in a beaker and heat them on burner as shown in the figure.
  2. We notice that ice slowly melts and becomes water and on further heating it changes to steam.
  3. If we reduce the temperature, the water vapour changes back to water and when temperature is further reduced it changes to ice.
  4. In the above activity we notice the change of the state of ice to water and to vapour but the substance, water, remains the same.
  5. Changes of this type where no new substance is formed are known as physical changes.
  6. When a material undergoes a change in shape, size, colour or state it is called a Physical Change.
  7. Generally, no new substance is formed in a physical change.

Activity – 4

Question 4.
Describe how do you perform the activity to observe the reaction of vinegar with baking soda?
Answer:
AP Board 7th Class Science Solutions 10th Lesson Changes Around Us 3

  1. First set up the apparatus as shown in Fig.
  2. Take a teaspoon of vinegar (acetic acid) in a test tube and add a pinch of baking soda (Sodium bicarbonate) to it.
  3. You observe bubbles coming out with a hissing sound. Pass this gas through freshly prepared Limewater (Calcium Hydroxide).
  4. Limewater changes to milky white showing that the gas sent into the test tube is Carbon dioxide.
  5. Vinegar + Baking Soda → Sodium acetate + Carbon dioxide + Water
  6. Carbon dioxide + Lime Water → Calcium Carbonate + Water
  7. In these reactions the new substances like Carbon dioxide and Calcium Carbonate are formed. Hence it is a chemical change.
  8. When a material undergoes a change in its composition it is called a chemical change.

AP Board 7th Class Science Solutions 10th Lesson Changes Around Us

Activity – 5

Question 5.
Write the approximate time after which they repeat.

NaturalPeriod of time of repetetion (Approximate)
1. Change of day and nightEvery 12 hours
2. Withering of leaves
3. Rising of the pole star
4. Change of Seasons
5. Appearance of full moon

Answer:

NaturalPeriod of time of repetetion (Approximate)
1. Change of day and nightEvery 12 hours
2. Withering of leaves1 year
3. Rising of the pole star24 hours
4. Change of Seasons1 year
5. Appearance of full moon30 days

Activity – 6

Question 6.
How do you produce large size of sugar crystals ? What type of change is it?
Answer:
AP Board 7th Class Science Solutions 10th Lesson Changes Around Us 4

  1. Take a big size test tube. Fill half of it with water. Add some sugar to it and stir it.
  2. Keep adding sugar and stirring until saturation is attained.
  3. Then heat this sugar solution and add some more sugar to it while stirring continuously.
  4. Continue adding sugar till no more sugar can be dissolved in it.
  5. Now filter the solution and allow it cool for half an hour.
  6. We notice formation of large size crystals of sugar at the bottom of the beaker. Thus the small granules of sugar added changed into large size sugar crystals.
  7. This is a physical change.

Activity – 7

Question 7.
Some changes are given in the table. Write possible changes you notice for each case and put (✓) in the appropriate column.
Answer:
AP Board 7th Class Science Solutions 10th Lesson Changes Around Us 5

1) In the above activity we notice that only in some examples like burning of paper, burning of crackers, change of milk to curd, boiling of egg, etc., a new substance is formed.

2) But in other examples of changes we notice an change in state or colour or size or shape etc but the substance remains same and no new substance is formed.

Activity – 9

Question 9.
Complete the following table.

Name of the FruitWhether turned brown or not
YesNo
1. Apple
2. Brinjal
3. Potato
4. Tomato
5. Cucumber
6. Mango

Answer:

Name of the FruitWhether turned brown or not
YesNo
1. Apple
2. Brinjal
3. Potato
4. Tomato
5. Cucumber
6. Mango

Some fruits and vegetables when cut, react with oxygen in the air. The process of reaction with oxygen is called oxidation. Due to this process brown layer is formed on the surface of fruits and vegetables.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

SCERT AP 7th Class Science Study Material Pdf 8th Lesson Wonders of Light Textbook Questions and Answers.

AP State Syllabus 7th Class Science 8th Lesson Questions and Answers Wonders of Light

7th Class Science 8th Lesson Wonders of Light Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Image formed by convex mirror is erect, small and _______
2. Infinite number of images are formed between two mirrors if they are kept at an angle of _______
3. Dentists use _______ mirror to see inner parts of teeth.
4. Image that cannot be obtained on screen is______
Answer:
1. virtual
2. 0°
3. concave
4. virtual

II. Choose the correct answer.

1. In periscope, the angle between two plane mirrors is
a) 0°
b) 30°
c) 45°
d) 60°
Answer:
a) 0°

2. If two plane mirrors are kept at an angle of 180° between them. Then the number of
a) 1
b) 2
c) 3
d) 4
Answer:
a) 1

3. Which of the following is not the property of image formed by a plane mirror?
a) Same size
b) Real
c) Lateral inversion
d) Erect
Answer:
b) Real

4. If a light ray falls on plane mirror along the direction of normal, then the angle of reflection will be
a) 90°
b) 45°
c) 0°
d) 180°
Answer:
a) 90°

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

5. Which form clear image among the following?
a) Paper
b) Cloth
c) Card board
d) Plane mirror
Answer:
d) Plane mirror

III. Match the following.

A) Reflections from Ice pieces1. Regular reflection
B) Reflections from still water2. Irregular reflection
C) Reflectors in head lights3. Blue light
D) Rear view mirrors4. Concave mirror
E) Light which causes damage to retina5. Yellow light
6. Convex mirror

Answer:

A) Reflections from Ice pieces2. Irregular reflection
B) Reflections from still water1. Regular reflection
C) Reflectors in head lights4. Concave mirror
D) Rear view mirrors6. Convex mirror
E) Light which causes damage to retina3. Blue light

IV. Answer the following questions.

Question 1.
What is reflection of light? Explain with an example.
Answer:

  1. The process of bouncing back of light rays into the same medium after falling on a smooth or rough surface from the light source is called “reflection of light”.
    Example:
  2. Take a plane mirror and stand in front of a building in open space during the day. Let the sunlight falls on the mirror. Now rotate the mirror in such a way that a spot of light falls on the wall of building.
  3. In the above activity, light rays fall on the mirror bounced back. This process of bouncing back of light rays in the same medium after hitting the surface of an object is called reflection.

Question 2.
Write the laws of reflection of light.
Answer:
There are two laws of reflection :

  1. First law of reflection : Angle of incidence is equal to angle of reflection.
  2. Second law of reflection : The incident ray, reflected ray and normal to the surface are present in the same plane. Incident and reflected rays are on either side of normal.

Question 3.
Why should we keep the mirror strips in the Periscope parallel to each other? What happens if they are not parallel?
Answer:

  1. We should keep the mirror strips in the Periscope parallel to each other. The image of an object formed on the first mirror reflects exactly on the second mirror and then we can see the image. It is possible when the mirror strips in the Periscope parallel to each other.
  2. If they are not parallel, the image of an object formed on the first mirror does not reflects on the second mirror and we cannot see the image through the second mirror.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Question 4.
What precautions do you take while getting the image of burning candle using a mirror?
Answer:

  1. Keep the candle as its flame is in the line of the mirror.
  2. Keep the mirror as its plane opposite to the screen.
  3. Do the experiment in a dark room.
  4. Take care while handling the candle with flame.

Question 5.
Write the properties of images formed by concave mirror.
Answer:
The properties of images formed by concave mirror.

  1. It can form both real and virtual images.
  2. It produces erect and inverted images
  3. It also produces small same size and bigger images depending on the position of object in front of it.

Question 6.
Draw the diagram show regular reflection and diffused reflection of light.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 1

Question 7.
Where do you find multiple images formed by plane mirrors in your day-to-day life?
Answer:

  1. In sweet shops mirrors are arranged in such a way that multiple images of sweets can appear.
  2. In saloon and beauty parlours mirrors are arranged in such a way that the customers can view his images in all directions.
  3. In bathrooms and dressing rooms also, we can observe multiple images.
  4. In Kaleidoscope we can observe multiple images.

Question 8.
Where do you find irregular reflections in your daily life?
Answer:
We can find irregular reflections on unpolished marbles floors, scratched steel vessels, broken mobile screens, disturbed water surfaces, and on the rough surfaces of plastic sheets, metal sheets, etc.

Question 9.
Write the applications of spherical mirrors in daily life.
Answer:
Applications of concave mirrors:

  1. Concave mirrors are used by ENT doctors as Head Mirrors.
  2. Dentists also use concave mirrors to get a bigger image of the teeth.
  3. Eye specialist using a special instrument called Ophthalmoscope. It is fitted with a concave mirror having a small hole near its center.
  4. Concave mirrors are used to reflectors in a torch light or the headlights of vehicles.
  5. These are used in microscopes.

Applications of convex mirrors:

  1. Convex mirrors are used as Rear-view mirrors in vehicles.
  2. Convex mirrors are also used at the junctions of roads.
  3. Convex mirrors are used at ATM machines.
  4. Convex mirrors are used for security purpose at factories, offices, etc.

7th Class Science 8th Lesson Wonders of Light InText Questions and Answers

7th Class Science Textbook Page No. 16

Question 1.
What do you observe in the picture? Mention the different objects in the scenery.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 24
Answer:
Trees, river, hills, rainbow, steps, boat, a boy, a girl, butterfly, water, plants, wooden bridge.

Question 2.
Can we see all these objects during night time?
Answer:
No, we cannot see all these objects during night time.

7th Class Science Textbook Page No. 17

Question 3.
Where does the light come from?
Answer:
Light comes from different objects called sources of light.
Ex : Sun, candle, tubelight.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Question 4.
Name some sources of light around you.
Answer:
Sun, candle, tubelight, bulb, torch light, match box etc .

Question 5.
Do these sources release light on their own?
Answer:
Sources like sun, stars emit light on their own.
Sources like bulb, torch light, candle do not emit light on their own.

Question 6.
Can you give some other examples for natural sources of light?
Answer:
The moon, stars and the sun.

Question 7.
Do you know how this light travel from the source?
Answer:
Actually, light is not a single ray, but a bundle of rays- This bundle of light rays are called beam of light rays. .

Question 8.
What do we call the path of light?
Answer:
The direction or path along which light travels is called a ray of light.

7th Class Science Textbook Page No. 18

Question 9.
How are we able to see the objects?
Answer:
The light fall on the objects and reflects and reach to our eye.

Question 10.
Is only light enough to see the objects around us?
Answer:
Yes, light is enough, but reflected light should reach our eye.

7th Class Science Textbook Page No. 19

Question 11.
In the activity-2, is there any reflection from the reflected light?
Answer:
Yes, there is reflection from the reflected light.

7th Class Science Textbook Page No. 23

Question 12.
We are able to see our image in the mirror. Can we catch our image bn screen?
Answer:
No. We cannot catch our image on screep.

7th Class Science Textbook Page No. 24

Question 13.
Are the images formed by plane mirrors real or virtual?
Answer:
Plane mirror forms virtual image.

Question 14.
Have you observed the arrangement of plane mirrors in a sweet shop? How many images did you find?
Answer:
Yes, I have observed the arrangement of plane mirrors in a sweet shop. So many images are formed.

7th Class Science Textbook Page No. 25

Question 15.
How many images are formed when the angle between two mirrors is zero?
Answer:
Number of images (n) = (360°/θ) – 1= ∞ , from the formula, we can find infinite number of images are formed. –

7th Class Science Textbook Page No. 26

Question 16.
Have you ever seen your image in rearview mirror of your father’s, motor cycle?
Answer:
Yes.

Question 17.
Is it same as the image seen in the mirror of your dressing table?
Answer:
No

Question 18.
What difference did you observe?
Answer:
Image is small in and closure.

Question 19.
What may be the reason for the difference?
Answer:
May be shape of the surface of the mirror.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Question 20.
Touch the surface of the rearview mirror. Is it same as plane mirror?
Answer:
No. It is not same as plane mirror. It is in curved in shape.

7th Class Science Textbook Page No. 28

Question 21.
Do you know the use of spherical mirrors in your daily life?
Answer:
Concave mirrors are used by dentists, ophthalmologists, ENT doctors and also in head lights of vehicles. Convex mirrors are used as rear view mirrors in vehicles and safety mirrors at curved roads.

7th Class Science Textbook Page No. 29

Question 22.
Have you ever observed the surface of reflection in a torch light or the headlights of vehicles?
Answer:
Yes, they are spherical in shape.

7th Class Science Textbook Page No. 30

Question 23.
Can you recognize the mirrors used beside the drivers in vehicles?
Answer:
These are convex mirrors.

Question 24.
Can you now recognize the characteristics of image from rear view mirror?
Answer:
Erect, virtual and smaller images are formed.

Question 25.
Do all the objects reflect the light rays falling on it?
Answer:
No.

Question 26.
Do you know why did paper burnt with a magnifying glass?
Answer:
The paper burned as the magnifying glass concentrated the sun’s rays at one place on the paper.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Question 27.
Doesthe magnifying glass reflect the light falling on it?
Answer:
No.

Think & Respond

7th Class Science Textbook Page No. 23

Question 1.
Deepak saw a vehicle on the road. He was surprised to see that the word AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 25 on it. What was that word? Why was it written in a strange manner?
Answer:

  1. The word is AMBULANCE.
  2. AMBULANCES are emergency vehicles.
    It is important to give way to ambulances.

The drivers can see the rear coming vehicles through the rear view mirror of the vehicles. Objects and letters appear in reverse means in left right inversion in mirrors. So letters in the wf>rd AMBULANCE also appear in reverse and cannot understand the word to the drivers.

If the word is written in reverse (as in mirror image) on the ambulance, it appears normally and readably. Drivers can read the word AMBULANCE and give a way to ambulance immediately.

Activities and Projects

Question 1.
Large sized concave mirrors are used to focus sun rays in solar cookers and solar heaters.Make a solar cooker of your own by using this principle and exhibit in your school science fair.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 2
Making of solar cooker :

  1. A concave mirror focuses parallel sun rays at the focal point of the mirror.
  2. So with a small concave mirror .we can heat up and burn paper.
  3. In the same way make a big concave mirror to heat up a vessel.
  4. Make a wooden/ iron frame in the shape of TV dish.
  5. Cut acrylic mirror sheets into 8 or 12 pieces in the shape of isosceles triangles with a height equal to the radius of dish antenna.
  6. The bases of 8 or 12 triangles together make the circumference of the dish.
  7. Stick the triangle mirrors to the dish.
  8. Our solar heater/cooker is ready.
  9. Arrange it so that concave part faces sun.
  10. Find its focal point and place a vessel at that point.
  11. It will get heated. We can even cook rice in that vessel.

Question 2.
Prepare a report by collecting data on the uses of spherical mirrors.
Answer:
Spherical mirrors are two types : They are 1) concave mirror 2) convex mirror.

1) Concave mirror :
Concave mirrors have a smooth curved surface inwards. Light rays are converged at one point by a Concave mirror. Real and virtual images can form the concave mirror.

The applications of concave mirror in our daily life :
Dentists :
Concave mirrors are used by dentists in the examination of teeth.

ENT doctors:
These are used by ENT doctors to examine the body parts.

Shaving mirrors :
During shaving, the concave mirror forms an enlarged and erect image of the face when the mirror is held closer to the face.

Optical instruments:
Ophthalmoscope concave mirrors are used in optical instruments such as Ophthalmoscope.

Astronomical telescopes:
These are used in telescopes and other devices. Headlights : Concave mirrors are widely used in headlights of automobiles and motor vehicles, torchlights, railway engines, etc. as reflectors. The light source is placed at the focus of the mirror, so after reflecting the light rays travel over a huge distance as parallel light beams of high intensity.

Solar cookers :
Concave mirrors are also used in solar cookers. Concave mirrors are also used in satellite dishes, electronic microscope, visual bomb detectors, etc.

2) Convex mirror .
Convex mirrors have a smooth curved surface outwards. Light rays are diverged at one point by a Convex mirror. Only virtual images can form the convex mirror.

The applications of convex mirror in our daily life.

Rear-view mirrors:
These are convex mirrors. These mirrors form images of objects spread over a large area. So, these help the driver to see the traffic behind them. Hence these are used as Rear-view mirrors.

Road safety:
Convey mirrors kept at the corners of curved roads to avoid accidents. Convex mirrors form images of vehicles on both sides of curved road. These mirrors are also used at the mnctions of roads.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Question 3.
Make a list of objects in your school and home, that works like a mirror. Make a report why they work like that.
Answer:

  1. Plane mirror : It has plane surface. So it reflects the light.
  2. Water surface : It has plane surface, when it was stable. So it reflects the light.
  3. Tiles of floor : It has plane surface. So it reflects the light.
  4. Almara : It was painted and has a smooth plane. So it reflects the light.
  5. Steel plate : It has a smooth plane. So it reflects the light.
  6. Concave and convex mirrors : It has plane surface. So it reflects the light.
  7. Computer screen : It has plane surface. So it reflects the light.
  8. Steel spoon : It has plane surface. So it reflects the light.

Question 4.
Take an empty tooth paste box and two mirror strips of required size and make a periscope.
Answer:
Collect the following materials to make your own periscope.

  1. Empty tooth paste box, two mirror strips, scale, pencil, blade, gum.
  2. Take an empty tooth paste box.
  3. Close the ends and draw two squares at both the ends by using scale and pencil.
  4. Draw diagonals to the square boxes, such that the two diagonals facing each other.
  5. Make two slits with the help of blade along the diagonals.
  6. Make sure that the size of mirror should be equal to the length of slits.
  7. Fix the mirror strips in the slits in such a way that reflecting surfaces of mirror strips face pach other.
  8. Fix the mirror strips firmly to the box by using gum.
  9. Cut out two windows on the narrow sides such that the windows should open directly on the reflecting surfaces of the mirror strips. Now the Periscope is ready.

Activities

Activity – 1

Question 1.
Write an activity to show the following type of light rays,
i) Parallel beam of light rays
ii) Converging beam of light rays
iii) Diverging beam of light rays
Answer:
i) Parallel beam of light rays:
Light rays which travel parallel to each other are called Parallel beam of light rays. To understand about parallel beam of light rays observe situation 1.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 3

Situation-1 :

  1. Take a plank and cardboard.
  2. Make small slits on cardboard.
  3. Keep the cardboard on the plank perpendicular to it.
  4. Keep it in sunlight during afternoon as shown in figure.
  5. The light rays from the sun fall on the cardboard and passed through the slits.
  6. We can observe that the light rays are travelling parallel to each other. These are called parallel beam of light rays.

ii) Converging beam of light rays :
“Light rays which travel from different directions to meet at a point are called as Converging beam of light rays”.
To know clearly about converging beam let us observe situation 2.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 4

Situation-2 :

  1. Recall the above situation.
  2. Arrange an acrylic mirror in the path of light rays as shown in the figure.
  3. The light rays meet at a point after bouncing back from the mirror. These are called converging beam of light rays.

iii) Diverging beam of light rays :
“Light rays which travel from a source moving in different directions are called as Diverging beam of light rays.” To understand about diverging beam, let us observe situation 3.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 5

Situation-3 :

  1. In the above situation, arrange the acrylic mirror in the path of light rays as shown in the figure.
  2. After bouncing back from the mirror light rays travel in different directions

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Activity – 2

Question 2.
Write an activity to Understand the reflection of light.
Answer:
“The process of bouncing back of light rays into the same medium after falling on a smooth or rough surface from the light source is called “reflection of light”. Let us understand this phenomenon with the following activity.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 6

  1. Take a torch and enter into a dark room.
  2. Observe the objects in darkness. We cannot see any object.
  3. Switch on the torch and focus it on the objects. We can see the objects.
  4. Keep a card board between your eyes and object.
  5. Now try to see the objects. We cannot see any object. ,
  6. From the above activity we can conclude that objects are visible only when light falls on the objects and bounces back to the eye.

Activity – 3

Question 3.
Write an activity to understand the reflection of light.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 7

  1. Take a plane mirror and stand infront of a building in open space during the day.
  2. Let the sunlight falls on the mirror.
  3. Now, rotate the mirror in such away that a spot of light falls on the wall of building.
  4. Light rays from the sun fall on the mirror and bounced back. The spot of Light on the wall is the image of sun.
  5. In this activity, light rays fall on the mirror and bounced back.
  6. This process of bouncing back of light rays in the same medium after hitting the surface of an object is called reflection.

Activity – 4

Question 4.
Write an activity to understand the regular reflection and irregular reflection of light.
Answer:
Reflection from a smooth and shiny surface is called regular reflection.

Reflection from an irregular or uneven surface is called irregular reflection or diffused reflection.

Let us understand these from the activity below.

1) Focus the light from a source on the objects mentioned in the given table and try to get the image of the objects.

2) Record your observations in the table given.

ObjectNature of surface of the object (Smooth and shiny/ Smooth but not shiny/ rough)Observation (Clear image is formed/image is formed but not clear/ No image is formed)
1. Plane mirrorSmooth and shinyClear image
2. New steel plateSmooth and shinyClear image/ but not clear
3. Card boardRoughNo image
4. Thermocol sheetRoughNo image
5. ClothRoughNo image
6. PaperRoughNo image

3) From the above table it is clear that light reflects regularly from a smooth and shiny surface and irregularly from an uneven surface.

4) Reflection from a smooth and shiny surface is called regular reflection.

5) Clear images are formed in case of regular reflection.

6) Reflection from an irregular or uneven surface is called irregular reflection or diffused reflection. Images are not clear or sometimes cannot form the images at all in case of irregular reflection.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Activity – 5

Question 5.
How can you verify the laws of reflection?
Answer:

  1. Take a White paper and spread it on a drawing board or cardboard.
  2. Draw a straight-line AB at the center of paper by a pencil.
  3. Mark a point O at the center of AB.
    AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 8
  4. At the point ‘O’ draw a perpendicular line ON to AB.
  5. ON is called normal to the surface.
  6. ake a protractor and place it on white paper in such a way that the zero edge of the protractor coincides with the point O on AB as shown in figure.
  7. Mark different angles starting from 10° to 80° on both sides of ON.
  8. Now place a plane mirror on the line AB with the help of two clips such that the surface of the plane mirror is perpendicular to the surface of the white paper.
  9. Take a laser pointer and focus it on plane mirror at ‘O’ making an angle of 40° with the line ON.
  10. Observe the reflected ray from the mirror.
    Angle of incidenceAngle of reflection
    1. 20°20°
    2. 40°40°
    3. 60°80°
    4. 80°80°
  11. The angle made by incident ray with the normal ON is called angle of incidence i.
  12. The angle made by the reflected ray with the normal is called the angle of reflection r.
  13. Focus laser pointer at different angles of incidence.
  14. Observe the angles of reflection in each case and tabulate the values.

From the above observations you can conclude that the angle of incidence is always equal to angle of reflection.
(OR)
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 10
This is the first law of reflection.
1) From the above activity you can also observe that Incident ray, reflected ray and normal are present in the same plane and incident, reflected rays are on either side of normal.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 9

2) The incident ray, reflected ray and normal to the surface are present in the same plane and incident, reflected rays are on either side of normal.
This is the second law of reflection.

Activity – 6

Question 6.
How can you verify the following characteristics of image by plane mirror?
1) Object distance is equal to image distance.
2) Size of the object is equal to size of the image.
Answer:
Let us verify the above characteristics by the following activity.
1) Take a chess board.
2) Fix a plane mirror vertically at the one end of the chess board as shown in the figure. Place a sharpener on one of the square boxes on the chess board.
3) You can see the image of sharpener in the mirror.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 11
4) Count the number of square boxes between the mirror and the sharpener. At the same time count the number of square boxes between the mirror and the image of sharpener. The number of square boxes equal in both the cases.
Conclusion 1:

5) From the above activity you can conclude that the distance of the object from the mirror is equal to the distance of image from the mirror.

6) The distance of the object from the mirror is called object distance. The distance of the image from the mirror is called image distance.

7) In the above activity observe the size of object [sharpener] and image.

8) You will find that the size of image and object are equal.

Conclusion 2:
From the above activity you can conclude that the size of the object is equal to size of the image.

Activity – 7

Question 7.
How can you verify the following characteristic of image by plane mirror?
Laterally inverted image is formed, (left and right alternates)
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 12

  1. Stand in infront of a plane mirror and raise your right hand.
  2. Observe the image formed in the mirror.
  3. Now lower the right hand and raise the left hand.
  4. Observe the changes you noticed in the image.
  5. Your right hand appears left and left hand appears right in the image.
  6. Such a shift of lateral side of images in opposite direction is called lateral inversion.

Conclusion:
So images formed by plane mirrors undergo lateral inversion

Activity – 8

Question 8.
How can you verify the following characteristics of image by plane mirror?
* The image formed is always virtual and erect.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 13

  1. Place a burning candle infront of the plane mirror.
  2. Observe the image of burning candle in the mirror.
  3. Now place a white paper behind the candle.
  4. Try to get the image of candle on paper.
  5. If your answer is ‘YES’, such type of image which we get on screen is called real image.
  6. If your answer is ‘NO’, such type of image which cannot get it on screen is called virtual image. •
  7. Plane mirror forms only virtual image.
  8. In the above activity, the image of burning candle is erected.
  9. We can see the erect image of the object in the mirror.

Conclusion:
Thus, we can conclude that plane mirrors form virtual and erect image.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light

Activity – 9

Question 9.
Write an activity to understand the relation between the arrangement of mirror and number of images.
(OR)
Derive a formula for number of images formed between two mirrors.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 14

  1. Take a drawing board, spread a white sheet on it.
  2. Draw a semicircle on the white sheet.
  3. Mark the angles from 0° to 180° with the help of protractor.
  4. Take two plane mirrors of same size join them with cellophane tape as shown in the figure.
  5. Keep the mirrors on the semicircle such that the angle between the mirrors is 120° .
  6. Now bring a burning candle in between the mirrors.
  7. Observe the number of images formed by the two Mirrors.
  8. Change the angles between the mirrors from 120° to 90°, 60°, 45°, and 30° respectively.
  9. Observe the number of images formed in each position. Tabulate your observation.

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 15
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 16
In this way we. can derive the formula to find out the number of images formed between two mirrors.

Number of images (n) = (360°/θ) -1 where 0 is the angle between the mirrors

Activity – 10

Question 10.

How do you make a Royal Road in a Shoe Box by using plane mirror?
(OR)
How do you prepare a Fancy item by using the phenomena of multiple reflections of plane mirrors?
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 17

  1. Fix two plane mirrors on the opposite sides of a shoe box with their reflecting surfaces facing each other.
  2. Draw a road on the bottom of shoe box between the mirrors.
  3. Arrange two LED street lights on both sides of road.
  4. Make a small hole on mirror side of shoe box and scrape off a little of colour coating from the mirror on that place.
    Observe the scenery inside the box through this hole.
  5. Light rays from the LED fall on the first mirror and then continuously undergone multiple reflections.
  6. That’s why we see a long road With full of LED lights.

Activity – 11

Question 11.
How do you make a periscope by using plane mirrors?
(OR)
Make your own periscope.
Answer:
Required materials:
Empty Incense stick box [Agarbatt(box], two mirror strips, scale, pencil, blade, gum.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 18

Procedure:

  1. Take an empty Incense stick box.
  2. Close the ends and draw two squares at both the ends by using scale and pencil.
  3. Draw diagonals to the square boxes, such that the two di-agonals facing each other.
  4. Make two slits with the help of blade along the diagonals.
  5. Make sure that the size of mirror should be equal to the length of slits.
  6. Fix the mirror strips in the slits in such a way that reflecting surfaces of mirror strips face each other.
  7. Fix the mirror strips firmly to the box by using gum.
  8. Cut out two windows on the narrow sides such that the windows should open directly on the reflecting surfaces of the mirror strips as shown in figure.
  9. Now your Periscope is ready.
  10. You can use this to see the objects outside the room through the window while hiding yourself in the room

AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 19

Activity – 12

Question 12.
Write an activity to understand different shapes of spherical mirrors.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 20
1) Take a stainless-steel serving spoon.
2) Look into it by bringing the outer bulged side of the spoon near your face.
3) Do you see your image in it? (Yes)
4) Did you observe any difference in image from what you see in plane mirror? (Yes)
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 21
10) Try to increase the distance of the spoon from your face.
11) Here the surface of the spoon act as a mirror but the surface is not plain.
12) The outer surface of the spoon is curved out like the outer surface of a sphere.
13) The inner surface of the spoon is like the part of inner surface of a sphere.
14) These mirrors are the parts of spheres. So these kinds of mirrors are called spherical mirrors.
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 22

Spherical mirrors are of two types :

  1. Convex mirrors (Reflecting surface bent outward)
  2. Concave mirrors (Reflecting surface bent inward)

Activity – 13

Question 13.
Characteristics of image formed by a concave mirror.
Take a ‘V’ shaped wooden stand. Keep a concave mirror on it. Place a lighted candle at a distance of about 50 cm in front of the mirror. Try to get clear image of the candle flame on a white screen or white paper by moving the screen towards and away from the mirror. Make sure that the screen is not placed in between burning candle and mirror.
AP Board 7th Class Science Solutions Chapter 8 కాంతితో అద్భుతాలు 22

Place the lighted candle at different distances in front of the mirror by moving it towards the mirror. Try to get a clear image on the screen at each position. Note your observation in the table given below.
Answer:
AP Board 7th Class Science Solutions 8th Lesson Wonders of Light 23
From the above table, you can conclude that a concave mirror can form real and virtual images, erect and inverted images, smaller, same size, and bigger images depending on the position of the object in front of it.

AP Board 7th Class Science Solutions 6th Lesson Electricity

SCERT AP 7th Class Science Study Material Pdf 6th Lesson Electricity Textbook Questions and Answers.

AP State Syllabus 7th Class Science 6th Lesson Questions and Answers Electricity

7th Class Science 6th Lesson Electricity Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Longer line in the symbol for a cell represents its ____ terminal and the shorter line represents ____ terminal.
2. Iron box works based on ____ effect of electric current.
3. The symbol for open switch is _______ .
4. The combination of two or more cells is called a ____
5. Expand MCB ____
Answer:
1. Positive, Negative
2. heat
3. AP Board 7th Class Science Solutions 6th Lesson Electricity 1
4. battery
5. Miniature Circuit Breaker

II. Choose the correct answer.

1. When cells are connected in series, it has
a) Single loop connection
b) Two loops
c) Many loops
d) None
Answer:
a) Single loop connection

2. Four bulbs are connected in parallel. When one bulb is removed then the remaining bulbs are
a) Put off
b) Continuously glowing
c) Blinking
d) None
Answer:
b) Continuously glowing

3. Statement 1: Generation of heat due to electric current is known Magnetic effect of electricity.
Statement 2 : An electromagnet works on magnetic effect of Electricity.
a) Both statements are true.
b) Both statements are false.
c) Statement 1 is true, Statement 2 is false.
d) Statement 1 is false, Statement 2 is true.
Answer:
d) Statement 1 is false, Statement 2 is true.

AP Board 7th Class Science Solutions 6th Lesson Electricity

4. Ramu’s house has five 60W bulbs. If they used all the bulbs for 5 hours. Then the power consumed by him
AP Board 7th Class Science Solutions 6th Lesson Electricity 2
a) 1500 KWH
b) 0.3 KWH
c) 70 KWH
d) 1.5 KWH
Answer:
d) 1.5 KWH

5. Which of the following works on the basis of heating effect of electric current?
a) Electric Crane
b) Electric Lift
c) Escalator
d) Hair drier
Answer:
d) Hair drier

III.Matching

A) MCB1. Safety device used in electric current
B) Fuse2. Two or more cells joined together
C) Battery3. Compact Fluorescent Lamp
D) Cell k4. Miniature Circuit Breaker
E) CFL5. Converts chemical energy into electrical energy
6. Useful for opening and closing circuit

Answer:

A) MCB4. Miniature Circuit Breaker
B) Fuse1. Safety device used in electric current
C) Battery2. Two or more cells joined together
D) Cell k5. Converts chemical energy into electrical energy
E) CFL3. Compact Fluorescent Lamp

IV. Answer the following questions.

Question 1.
Why are bulbs in series put off when a bulb is removed from it?
Answer:

  1. In series connection electricity has only one path.
  2. If a bulb is removed from the series connection of bulbs, the circuit becomes open.
  3. So, remaining bulbs are put off as the circuit open.

Question 2.
Read the following riddles and answer them.
1. I am used to close-or open the circuit. Who am I?
2. I will give you light. Who am I?
3. I convert chemical energy into electrical energy. Who am I?
4. Without me you cannot connect the electrical components. Who am I?
Answer:

  1. Switch
  2. Bulb
  3. Cell
  4. Connecting wires

Question 3.
Give examples for electric appliances which work on the property of heat effect of electricity.
Answer:
Examples for electric appliances which work on the property of heat effect of electricity:

  1. Electric heater
  2. Electric iron box
  3. Soldering gun
  4. Electric kettle
  5. Electric rice cooker

AP Board 7th Class Science Solutions 6th Lesson Electricity

Question 4.
A cell, 5 bulbs and a switch are connected in a circuit but bulbs are not glowing. Write the possible reasons for this.
Answer:
Possible reasons:

  1. May be connected a dysfunctional or less voltage cell.
  2. May be switch in off position.
  3. Wires may be cut any where or not properly connected.
  4. At least one of the bulbs may be damaged (If they are connected in series).
  5. All the bulbs may be damaged. (If they are connected in parallel)

Question 5.
Describe the process to prepare an electromagnet.
Answer:
1) Take an iron nail.
2) Wound the insulated copper wire tightly around the nail.
3) Connect the two ends of the coiled copper wire to a battery and a switch in series connection.
4) AP Board 7th Class Science Solutions 6th Lesson Electricity 3
5) Place some pins near the iron nail.
6) Now switch on the circuit.
7) Observe that all pins cling to the end of the iron nail.
8) This indicates coiled copper wire around nail acts as magnet when electricity passes through it.
9) It means the coiled copper wire around the nail acts as an electromagnet.
10) Now switch off the circuit.
11) Observe that all pins fall from the iron nail.
12) Hence, we can say that the coiled copper wire does not act as a magnet when no electricity passes through the circuit.

Question 6.
Draw the symbols for the following.
A) Bulb
B) Cell
C) Battery
D) Open Switch
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 3

Question 7.
Draw a circuit diagram of electrical components in a sequence by using cell, bulb, switch is off mode and connecting wires.
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 4

Question 8.
How do you appreciate the role of magnetic effect and heating effect in our daily life?
Answer:

  1. The production of heat due to the flow of electricity through a wire is called heating effect of electricity.
  2. We are using many home appliances like electric iron boxes, electric cookers, electric heaters, geysers, hair dryers which work on the principle of heating effects of electricity.
  3. The above said home appliances contain heating elements which produce heat.
  4. The production of magnetic force due to the flow of electricity through a wire is called magnetic effect of electricity.
  5. Loud speakers, MRI machines, generators, metal detectors, mobile phones, mixer grinders, electric fans, electric bells, electric motors etc.; work based on the magnetic effect of electricity.
  6. In this, way, heating and magnetic, effects of electricity are used in medical, automobile, industrial, agricultural etc., fields and also in homes.
  7. The devices and home appliances have made our lives more comfortable.
  8. Hence, I appreciate the role of magnetic effect and heating effect^ in our daily life.

Question 9.
Prepare some slogans to create awareness on wastage of electricity.
Answer:

  1. Save electricity – Save money.
  2. Turn off the lights, before leaving.
  3. Today’s wastage is tomorrow’s shortage.
  4. The less you burn, the more you earn.
  5. Save power – Save nation.
  6. Be polite – turn off the light.
  7. When the sun is bright – say no to tube light.
  8. Energy is precious – use 5 star appliances.
  9. Leave filament bulbs – Use LED’S
  10. Less consumption of power – Leads more comforts in future.

Question 10.
Suggest first aid when a person suffers electric shock.
Answer:
First aid for electric shock :

  1. If the shock is severe and the person is unconscious, give mouth to mouth artificial respiration.
  2. Sometimes heart beat ceases, at that time keep your palms on his chest and press down and release until the heart beat is recovered. This is called Cardio Pulmonary Resuscitation (CPR).
  3. Immediately take him to the hospital.

7th Class Science 6th Lesson Electricity InText Questions and Answers

7th Class Science Textbook Page No. 83

Observe the picture carefully.
AP Board 7th Class Science Solutions 6th Lesson Electricity 12

Question 1.
What are the electric devices you find in the above picture?
Answer:
1) Bulb 2) Fan 3) TV.

Question 2.
List out the electric devices used in your homes.
Answer:
1) TV 2) Fans 3) CFL bulbs 4) LED bulbs 5) Tube light (LED) 6) Music system 7) Chargers 8) Refrigerator 9) Electric geyser 10) Vapourizer 11) Iron box 12) Calling bell 13) Mixie 14) Wet grinder 15) AC 16) Motor pump 17) Mosquito repellent.

7th Class Science Textbook Page No. 84

Question 3.
Why does a bulb glow when we switch it ON?
Answer:
When we switched ON the bulb electricity flows through the wire and makes the bulb to glow.

Question 4.
What do you observe inside the cell?
Answer:

  1. Carbon rod with a metal cap.
  2. Carbon powder.
  3. Ammonium chloride paste.

Question 5.
Can we make a cell of our own?
Answer:
Yes, we can make a cell by using a zinc plate, copper plate, a small bulb, connecting wires, lemon or orange, crocodile clips.

7th Class Science Textbook Page No. 85

AP Board 7th Class Science Solutions 6th Lesson Electricity 13
Question 6.
Do you know the type of cell is used in wrist watch?
Answer:
Button cells are used is wrist watches.

Question 7.
How many types of electric cells shown above are known to you?
Answer:
I know the dry cells, lithium cells, button cells and alkaline cells.

Question 8.
Where do we used dry cells?
Answer:
In our daily life we use dry cells in torch lights, wall clocks, radios etc.

7th Class Science Textbook Page No. 86

Question 9.
Does a bulb glow when switch is in off position? Why?
Answer:
When the switch is in off position, bulb does not glow. Because the circuit is opened.

7th Class Science Textbook Page No. 87

Question 10.
Among all these bulbs which bulb consumes less electricity?
Answer:

  1. We can observe star symbols (power saving star rating) are marked on electrical appliances.
  2. The number of stars on them indicates the energy efficiency of that device.
  3. The electrical appliance with more stars consumes less electricity.
  4. LED bulbs are more energy savers than other bulbs.

AP Board 7th Class Science Solutions 6th Lesson Electricity

Question 11.
Is there any protection provided for electric devices in our house?
Answer:
Yes. The fuse is used to protect the electric devices in our house.

Question 12.
If there is a power failure only in your house. What do you check first?
Answer:
We will check the fuse first.

Question 13.
Have you seen the electric device shown in the figure. Where is it used?
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 14

  1. The electrical devices shown in the figure are fuses.
  2. Fuses are used at meter board to protect the devices from excess flow of electricity.

Question 14.
Is there any other safety device which can be used as a fuse?
Answer:
Yes, MCB can be used as a fuse.

7th Class Science Textbook Page No. 88

AP Board 7th Class Science Solutions 6th Lesson Electricity 15
Question 15.
Have you ever seen this component?
Answer:
Yes, I saw it. It is a MCB.

Question 16.
Which is a better safety device, electric fuse or MCB?
Answer:
MCB is better, when compared to electric fuse. There is no need to change it often. But wire in electric fuse has to be changed whenever it melts.

Question 17.
Do you know how to show the connections of the electric components in a circuit?
Answer:
Yes. We can show the connections of the electric components in a circuit by its symbols.

Question 18.
What electric components do you find in the figure?
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 16
1) Switch 2) Bulb 3) Wire 4) Battery.

Question 19.
Is it possible to draw realistic electric components every time? Why?
Answer:

  1. Drawing circuit diagrams with realistic components is difficult.
  2. If the circuit is a bigger one with several components it is much more difficult.
  3. To overcome this, standardized symbols of electric components are used to draw the circuit diagrams.

7th Class Science Textbook Page No. 89

Question 20.
Isn’t it so easy to draw like this?
Answer:
It is easy to draw a circuit by using symbols of electric components than realistic components. .

Question 21.
What difference do you find between circuit diagrams A and B?
AP Board 7th Class Science Solutions 6th Lesson Electricity 17
Answer:

  1. We observe that the bulb in circuit ‘A’ is not glowing and in circuit ‘B’ it is glowing.
  2. We can also see the switch is in OFF mode in circuit diagram A and it is in ON mode in circuit diagram B.
  3. Circuit diagram ‘A’ represents open circuit, where as circuit diagram ’B’ represents closed circuit.

7th Class Science Textbook Page No. 90

Question 22.
Is it possible to connect more than one bulb or one cell in an electric circuit?
Answer:
Yes, it is possible.

Question 23.
What happens if we connect more bulbs or cells in an electric circuit?
Answer:

  1. There are two ways to connect more bulbs or cells in an electric circuit they are parallel and series connections.
  2. In parallel connection electricity has more than one path where as in series connection only one path.
  3. If the number of cells are connected in series the brightness of the bulbs enhanced where as in series the brightness of bulb remains same.

Question 24.
Have you observed a large number of bulbs used in decorations of marriage functions, festivals etc.?
Answer:

  1. Yes, I have observed a large number of bulbs used in decorations of marriage functions, festivals etc.
  2. All the bulbs are connected one after one that means in series.

AP Board 7th Class Science Solutions 6th Lesson Electricity

Question 25.
What happens when we connect two cells and two bulbs in the same path?
Answer:

  1. If we connect two cells and two bulbs in the same path, it is called series connection.
  2. The brightness of the bulbs increased.

7th Class Science Textbook Page No. 92

Question 26.
Where can we observe these type of connections? (Series & Parallel)
Answer:

  1. We can use series connection of cells in torch light, toys, remotes, wall clocks,radio etc.
  2. We can use parallel connection of cells for a long – lasting battery life.

Question 27.
Do you know how several electric devices are connected in your home? Did you observe any pattern in their connections?
Answer:
If All the electric devices in our house are connected in parallel.

7th Class Science Textbook Page No. 93

Question 28.
Where can we observe these types of connection?
Answer:
We use series connection of bulbs for decoration purpose, parallel connection of bulbs in house hold connections.

7th Class Science Textbook Page No. 94

Question 29.
Naime the electrical devices used to produce heat?
Answer:
Electric iron box, electric heater, geyser, grills, electric cooker etc. are the electrical devices used to produced heat.

Question 30.
Have you ever used an iron box to iron your clothes? How does it produce heat?
Answer:

  1. Electric iron box contains the coil of wire (filament) which produce heat, when electricity passed through it.
  2. It is called heating effect of electricity.

AP Board 7th Class Science Solutions 6th Lesson Electricity

Question 31.
Which electric appliances work on the principle of heating effect of electricity?
Answer:
1) Electric stove, 2) Electric room heater, 3) Electric iron box, 4) Electric kettle, 5) Electric cooker, 6) Water heater, 7) Geyser, 8) Hair dryer, etc. are working on the principle of heating effect of electricity.

7th Class Science Textbook Page No. 95

Question 32.
1) How does a fan work?
2) What makes the fan to rotate?
3) What is the principle involved in it?
Answer:

  1. Fan works on the principle of magnetic effect of electricity.
  2. The magnetic force, which is produced due to the flow of electricity through a wire (coil) makes the fah to rotate.
  3. The principle, magnetic effect of electricity is involved in it.

7th Class Science Textbook Page No. 96

Question 33.
Which electric appliances work on the principle of magnetic effect of electricity?
Answer:
1) Electric fan, 2) Electric bell, 3) Electric motor, 4) Loud speaker, 5) MRI machines, 6) Generators, 7) Metal detectors, 8) Mobile phones, 9) Mixer grinders, etc. work on the principle of magnetic effect of electricity.

Question 34.
Have you seen the electric meter in your house? What do you find in the meter?
Answer:
I find reading of consumed electricity in units.

Question 35.
What does the term ‘one unit’ mean?
Answer:
1 unit = 1000 watt = 1 K.W.H.

Question 36.
How is the usage of electricity measured?
Answer:
Usage of electricity is measured in KWH (unit).

7th Class Science Textbook Page No. 98

Question 37.
What is an electric shock and when does it occur?
Answer:

  1. An electric shock occurs when a person comes into contact with an electrical source.
  2. Electricity flows through a portion of the body causing a shock.

Think & Respond

7th Class Science Textbook Page No. 85

Question 1.
Now a days we are using rechargeable torchlights, bulbs, mobile phones and fens. Think and discuss with your classmates about the working of these electric devices and the cells used in diem.
Answer:

  1. A cell is made up of three parts, a cathode, an anode and an electrolyte (a chemical).
  2. A cell works by connecting its stored chemical energy, into electrical power
  3. One the electrolyte of the cell is used up it doesn’t work further in non – rechargeable cells (Alkaline cells).
  4. Electrons (negative charges) travel from an anode to a cathode until the anode is considered empty or dead.
  5. A rechargeable cell, a charger can reverse the electron flow again to restore at the anode. This is called charging.
  6. This reversing process depends upon the electrolyte used in the cell.
  7. Normal cells do not contain such type of electrolytes to reverse the charge.
  8. But, the cells like NiMH (Nickel – Metal Hydride), Lithium – ion are capable of reversing the charge when charged it for number of times.
  9. This type of rechargeable batteries are used in rechargeable torch lights, bulbs, mobile phones and fans.

7th Class Science Textbook Page No. 92

Question 2.
a) What happens if more cells are connected in a series?
b) Is there any restriction on the number of cells in a circuit for a given bulb?
Answer:
a) If more cells are connected in a series, voltage increases in the circuit. That means the strength of the current in the circuit increases.

b) When we will join lot of cells in a circuit, then the bulb will not be able to handle the large amount of energy supplied to it. Hence, the bulb may damage.

7th Class Science Textbook Page No. 95

Question 3.
Do all home appliances produce heat when electricity passes through them?
Answer:

  1. Yes, all home appliances produce heat when electricity passes through them.
  2. But the amount of heat produced depends on the material of conductor, such as length, thickness and nature of it.

7th Class Science Textbook Page No. 96

Question 4.
Is it possible to create permanent magnets using the magnetic property of electric current? Think and discuss with your classmates and make notes on this.
Answer:

  1. Electromagnets are not permanent magnets because they lose the property of magnetism when the circuit is open.
  2. A permanent magnet has a magnetic field all the time unless it is heated.
  3. But electromagnet doesn’t retain its magnetism all the time, without current.

AP Board 7th Class Science Solutions 6th Lesson Electricity

7th Class Science Textbook Page No. 98

Question 6.
Why don’t we use wet stick to push away a person when he get electric shock? Think and discuss with your friends.
Answer:

  1. Wet stick contains some amount of water throughout it.
  2. Water is a good conductor of electricity (ofcourse, pure water does not pass current).
  3. If we touch the person to push away from the electric wire, the current passes through the water into our body.
  4. It causes to get electric shock to us also.
  5. So, we should not use wet stick to push away a person when he get electric shock.
  6. We should use dry stick only to do that.

Activities and Projects

Question 1.
Collect any manual of electric equipment. Read the information in detail and answer the following questions.
A) When was it manufactured and how long can we use it?
B) How many stars are given to it?
C) Is it working on the basis of magnetic or heating effect?
Answer:
I Read the manual of a washing machine.
A) It was manufactured on 8<sup>th</sup> March 2020.
B) It was given 5 stars.
C) It is working on the bases of both magnetic effect (motor) and heating effect (in built heater).

Question 2.
Find out the electric meter readings for 3 months in your house. Record your observations. Ask your teacher about how electricity bill is paid.
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 5

Question 3.
Make a pamphlet on “Save electricity, don’t waste it”. Display it on the wall magazine?
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 6

Question 4.
Collect last month’s electricity bills from your friends or neighbours houses and record the data. Advise them to change their ordinary bulbs with CFL bulbs. Again collect the electricity bilis of next month and observe the differences between the two bills. Discuss your observations with your friends and suggest better bulb.
Answer:
Observation sheet (Student Activity).
AP Board 7th Class Science Solutions 6th Lesson Electricity 7
Conclusion :
After replacing ordinary bulbs with CFL/LED bulbs the consumption of electricity is reduced and Bill amount is saved.

Question 5.
If anyone around you has suffered an electric shock, collect the information along with their feelings and know about CPR. Record your observations in your note book and discuss with your friends.
Answer:
Observation sheet (Student Activity).
AP Board 7th Class Science Solutions 6th Lesson Electricity 8

Activities

Activity – 1

Question 1.
If you open a dry cell, what do you observe?
(OR)
What do you observe inside the cell?
(OR)
Explain the structure of a cell with an activity.
(OR)
Describe the constru-ction of a dry cell.
(OR)
Describe the various components in a dry cell and their functions.
Answer:
AP Board 7th Class Science Solutions 6th Lesson Electricity 9

  1. Take a cell used in torch light and open it with the help of your teacher.
  2. The cell consists of a container made of zinc metal. The container works as the negative terminal (- ve).
  3. A carbon rod with a metal cap works as a positive terminal 0 ve>
  4. The positive and negative terminals are called electrodes.
  5. The carbon rod is surrounded by a mixture of carbon powder and a chemical called ammonium chloride. This mixture acts as an electrolyte.
  6. The cell is sealed from the top.

Activity – 2

Question 2.
How do you make an own cell?
(OR)
Can you make a cell? If you can write the procedure.
(OR)
Write the procedure of making a cell using lemon.
Answer:
Material required: Zinc plate, Copper Plate, A small bulb or LED, Connecting wires and Lemon or Orange, crocodile clips-4.

Procedure:
Take a Lemon and insert a Zinc plate and a Copper plate into it as shown in the fig. These plates act as electrodes and the juice inside the fruit acts as electrolyte. Connect two terminals of the bulb to zinc and copper plates using connecting wires.
AP Board 7th Class Science Solutions 6th Lesson Electricity

We observe that the bulb will glow as electricity was produced using chemical energy present in the lemon.

Activity – 3

Question 3.
Write an activity two observe the brightness of bulb in the following situations.
1) One cell and one bulb.
2) Two cells in series and one bulb.
3) Two cells in parallel and one bulb.
(OR)
How is the brightness of a bulb when it connected to a circuit with cells which are connected i) in series and ii) in parallel? Explain with an activity.
Answer:
Situation-1
Material required :
Dry cell -1, Small bulb-1 (torch light bulb or LED), Switch and Connecting wire.
AP Board 7th Class Science Solutions 6th Lesson Electricity 4

Procedure:
Take a dry cell, small bulb or LED and a switch. Connect them in a simple circuit using copper wire as shown in given fig. Switch ON and observe the brightness of light. ,

Situation-2 (Cells connected in series) :
Material required :
Dry cells -2, Small bulb-1 (torch light bulb or LED), Switch and Copper wire.
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 9

Procedure:
Take two dry cells, small bulb and a switch. Connect them using a copper wire as shown in the fig. Switch ON and observe the brightness of light. Remove one cell and observe the condition of bulb.

Situation-3 (Cells connected in parallel) :
AP Board 7th Class Science Solutions Chapter 6 విద్యుత్ 10
Material required :
Dry cells -2, Small bulb -1 (torch light bulb or LED), Switch and Copper wire.

Procedure:
Take two electric cells small bulb or LED and switch. Connect them using copper wire as shown in fig.

Switch ON and observe the brightness of bulb.Remove one cell and observe the condition of bulb.Record your observations in the above three situations in the given table.
AP Board 7th Class Science Solutions 6th Lesson Electricity 10

From the table, if a number of cells are connected in series the brightness of the bulb isenhanced. If one of the cells is removed, the circuit becomes open and the bulb will not glow.If a number of cells are connected in parallel the brightness of the bulb remains same. Evenlf one of the cells is removed, the circuit remains same and the bulb will glow as usual.

Activity – 4

Question 4.
Write an activity to show the brightness of the bulbs when two bulbs are connected in series and in parallel.
Answer:
Situation-1
Materials required:
Dry cell -1, Small bulb-1 (torch light bulb or LED), Switch and Connecting wire.

Procedure :
Take a dry cell, small bulb or LED and switch.Connect them in a simple circuit using copper wireas shown fig.. Switch ON and observe the brightness of light.
AP Board 7th Class Science Solutions 6th Lesson Electricity 4

Situation-2 (Bulbs connected in series):
Materials required:
Electric cell -1, Small bulbs -2 (torch light bulb or LED), Switch and Copper wire.

Procedure :
Take two small bulbs or LED’s, cell and switch.
AP Board 7th Class Science Solutions Chapter 6 Electricity

Connect them using copper wire as shown in fig. Switch ON and observe the brightness of the bulbs. Remove one bulb and observe the condition of another bulb.

Situation-3 (Bulbs connected in Parallel):
Materials required:
Electric cell -1, small bulb -2 (torch light), Switch, Copper wire.
AP Board 7th Class Science Solutions Chapter 6 Electricity

Procedure:
Take two small bulbs, a cell and a switch.Connect them using copper wire as shown in fig. Switch ON and observe the brightness of the bulbs.

Remove one bulb and observe the condition of another bulb.
AP Board 7th Class Science Solutions 6th Lesson Electricity 11

From the table,if a number of bulbs are connected in series the brightness of the bulb is reduced. If one of the bulbs is removed or broken, the circuit becomes open and the second bulb will not glow. If the number of bulbs are connected in parallel the brightness of the bulb remains same. Even if one of the bulb Is removed or broken, the circuit remains same and the bulb will glow as usual, but the battery drains faster.

Activity – 5

Question 5.
Write an activity to know about beating effect of electric current.
(OR)
What happens if Garment is passed through a Nichrome wire? Explain with an activity.
(OR)
How do von show the heating effect of electricity through an activity?
Answer:
Materials required:
Electric cell, switch. Iron nails-2, wooden board, connecting wires, 10 cm of Nichrome wire.

Procedure :
Construct a circuit using electric cell, switch, connecting wires and iron nails in series connection. Insert the iron nails in the wooden board. Keep the switch in OFF position. Take Nichrome wire and tie it between the iron nails as shown in fig.
AP Board 7th Class Science Solutions Chapter 6 Electricity

Observations :
Now switch ON the circuit for one minute, and switch it OFF. Now, touch the Nichrome wire, (don’t hold Nichrome wire for a long time.) You will observe that the Nichrome wire gets heated up when electricity passed through it. The production of heat due to the flow7 of electricity through a wire is called Heating Effect of Electricity.

AP Board 7th Class Science Solutions 6th Lesson Electricity

Activity – 6

Question 6.
Write an activity to know about magnetic effect of electric current.
(OR)
How do you make an electromagnet?
(OR)
How do you prepare an electromagnet?
Answer:
Material required:
Battery, Switch Iron nail (nearly 8 cm). Insulated copper wire (50 cm) and small pins.

Procedure:
Take an iron nail and wound the insulated copper wire tightly around the nail. This iron aaS works as a coil. The two ends of the coiled copper wire are connected to a battery and a switch in series connection as shown in fig. (Switch roast be in OFF position. Place some pins near the iron nail and switch ON the circuit.

We observe that, when we switched, on the circuit, all pins ding to the end of the iron nail. This indicates coiled copper wire around nail acts as magnet when electricity passes through it. It means the coiled copper wire around the nail acts as an electromagnet.
AP Board 7th Class Science Solutions 6th Lesson Electricity 3

When we switched off the circuit, immediately all the pins fall from the iron naS. This indicates that coiled copper wire does not act as a magnet when no electricity7 passes through the circuit

Magnetic effects produced in a coil due to the passing of electricity in the coil is called magnetic effects of electricity.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

SCERT AP 7th Class Science Study Material Pdf 5th Lesson Motion and Time Textbook Questions and Answers.

AP State Syllabus 7th Class Science 5th Lesson Questions and Answers Motion and Time

7th Class Science 5th Lesson Motion and Time Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Units of speed is _____
2. If a body covers equal distances in equal intervals of time, it is said to be _____ motion.
3. 60 minutes = _____ seconds. (3600)
4. Average speed = total distance travelled / _____
5. ISRO stands for _____
6. Geostationary satellites are used for _____ purpose.
Answer:
1. m/s
2. uniform
3. total time taken to travel
4. Indian Space Research Organization
5. Communication

II. Choose the correct answer.

1. The correct relation between speed, distance and time is
a) Speed = distance / time
b) Speed = time / distance
c) Time = speed/ distance
d) Distance = speed / time
Answer:
a) Speed = distance / time

2. What is common in the following?
Motion of the propeller of a flying helicopter, the hands of a watch, motiob of a giant-wheel.
a) All are examples of translatory motion
b) All are examples of oscillatory motion
c) All are examples of rotatory motion
d) a & c
Answer:
c) All are examples of rotatory motion

3. Which of the following is not an oscillatory motion?
a) Motion of the hammer of an electric bell.
b) Motion of your hands while running.
c) Motion of a child on a see – saw.
d) Motion of a horse pulling a cart.
Answer:
d) Motion of a horse pulling a cart.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

4. Which of the following statements is not correct?
a) The basic unit of time is second.
b) Every object’s motion is Uniform.
c) 1 km/h = 5/18 m/s.
d) Speed is expressed in km/h.
Answer:
b) Every object’s motion is Uniform.

III. Matching

A) Rotatory Motion1. Rocket
B) Oscillatory Motion2. A train entering into a railway station
C) Translatory Motion3. Movement of needle in a sewing machine.
D) 100 years4. Decade
E) 10 years5. Movement of hands in a wrist watch.
6. Century

Answer:

A) Rotatory Motion5. Movement of hands in a wrist watch.
B) Oscillatory Motion3. Movement of needle in a sewing machine.
C) Translatory Motion2. A train entering into a railway station
D) 100 years6. Century
E) 10 years4. Decade

IV. Answer the following

Question 1.
The distance between two stations is 240km. A train takes 4 hrs to cover this distance. Calculate the speed of the train m/s.
Answer:
The distance between two stations = 240 km
Time taken to cover this distance = 4 hrs
Speed of the train = distance / time
= 240/ 4 km/hrs
= 60 km/hrs

This is the speed of the train in km/hrs,
to convert it into m/s….
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 1

Question 2.
Can an object possess translatory and rotatory motion at the same time? Give an example.
Answer:

  1. Yes. An object can possess both translatory and rotatory motions at the same time.
  2. When we observe the motion of a cycle tyre moving in a straight line, all the parts of tyre moves in the same direction of cycle so, we can say it is in translatory motion.
  3. If we observe the rolling of cycle tyre, it move in a curved (circular) path about a iron rod in the middle of the wheel as an axis. So, we can say it is in rotatory motion.
  4. In this way the cycle tyre moving in a straight line possesses both translatory, and rotatory motions.

Question 3.
In a sewing machine used by tailors, mention the type of motion of its parts when it runs, a) the wheel b) the needle c) the cloth
Answer:

Parts of sewing machineType of motion
The wheelRotatory motion
The needleOscillatory or vibratory motion
The clothTranslatory motion

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 4.
Write the motion of different parts of a bicycle while it is in motion, a) the wheels b) the cycle chains c) the pedal with its arm d) the movement of the rider along with the bicycle.
Answer:

Parts of a bicycleType of motion
The wheelsRotatory and translatory motions
The cycle chainsTranslatory motion (curvilinear motion in elliptical path)
The pedal with its armRotatory motion
The movement of the rider along with the bicycleTranslatory motion

Question 5.
John tied a stone to a string and whirled it around. What type of motion do you observe?
Answer:
When a stone is tied to a string and whirled it around, it move in a circular path around the hand. That means the stone fallows a curved path with respect to the hand as a fixed centre or axis of rotation. So it is said to be in rotatory motion.

Question 6.
What are Uniform and Non-Uniform motions? Give four examples for each type of Motions.
Answer:
i) Uniform motion : “If a body covers equal distances in equal intervals of time, it is said to be in uniform motion”.
Ex: 1. Hands of a clock
2. Revolution of Earth
3. A train going at a steady speed
4. The blades of a fan

ii) Non-uniform motion: If a body covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.
Ex : 1. Movements of butterfly in a garden
2. The motion of a car in a city traffic
3. An aircraft while landing
4. Running of athletes in a marathon

Question 7.
Your friend told you the motion of rocket is fast motion. Will you agree? How can you decide whether the motion of a body is slow or fast?
Answer:

  1. Yes. I agree with him. Because rockets move with the speed of 12 km per second. Its very high speed.
  2. The distance travelled by an object in a given interval of time can help us to decide whether it is faster or slower.
  3. The distance travelled by an object in a unit time is called as its Speed.
  4. Speed = Distance travelled / Time taken to travel.
  5. Units of speed is meter per second (m/s) or kilometre per hour (Km/h).
  6. 1 Km / h = 5/18 m/s
  7. The object moving with less speed is slow and more speed is fast. Thus we can decide whether the motion of a body is slow or fast using its speed.
  8. At a particular instant of time, we can find speed of a vehicle using speedometer.

Question 8.
How can you appreciate the clocks and watches in measuring or estimating time?
Answer:

  1. Clocks and watches are helpful to us in measuring or estimating time.
  2. They help us in scheduling our works properly.
  3. They help us in attending the works on time by displaying the time.
  4. With out clocks and watches its very difficult to estimate the time.
  5. Sometimes they are used to make a loud alert signal at a specified time to make us wake up or recall a work.
  6. Some times they are helpful in measuring the time also.
  7. Now a days they are.available in some other gadgets like mobile phone also.
  8. Thus they become an essential and integral part of our life.

7th Class Science 5th Lesson Motion and Time InText Questions and Answers

7th Class Science Textbook Page No. 68

Question 1.
What do you observe from the picture?
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 6
Answer:
You can see the movement of cars, buses, train, boat, clouds, birds and minute hand in the clock etc. On the other side we can see the bridge, buildings, road, hill and the clock tower which are not moving.

7th Class Science Textbook Page No. 69

Question 2.
Do all these things move alike?
Answer:
No.

Question 3.
What makes the objects move and moving object to come to rest?
Answer:
Force makes the objects move and moving object to come to rest.

Question 4.
What do we call the motion of a train moving in a straight line?
Answer:
Translatory motion.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 5.
Is the forward movement of birds and the movement of its wings same?
Answer:

  1. Not same.
  2. Forward movement of birds is a translatory motion and wing movement is an oscillatory motion.

Question 6.
How do you decide whether any object is in motion or at rest?
Answer:

  1. Motion is a common experience in our life.
  2. We feel an object is in rest when it is not changing its position and we fee! an object is in motion when it is changing its position with respect to the surroundings.

Question 7.
Are all these trees, buildings, electric poles, etc. really in motion?
Answer:
No.

Question 8.
Which one do you feel is moving, passenger in the bus or tree?
Answer:
Passenger.

Question 9.
Which one do you feel is moving, tree or your co-passenger?
Answer:
Tree.

Question 10.
Why is it so?
Answer:
Motion and rest are relative terms and ‘hey are relative to the observer. We feel an object is at rest when it is not changing, its position. We feel an object is in motion When it is changing its position with respect to the surroundings.

7th Class Science Textbook Page No. 70

Question 11.
Have you ever played carrom board? What is the reason for movement of coins when we hit them with a striker?
Answer:
The force applied by the striker on the coins is the reason for movement of coins.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 12.
How force effects the motion?
Answer:
Force can change the position of motion to rest, of vice – versa.

7th Class Science Textbook Page No. 71

Question 13.
What did your friend apply to make the ball move?
Answer:
Force (muscular force).

Question 14.
What did you apply to change the direction of the ball?
Answer:
Force (muscular force).

Question 15.
What did your friend apply to stop the moving ball?
Answer:
Force (muscular force).

Question 16.
What can we call the length of path that an object moves from one place to another?
Answer:
Distance.

Question 17.
Ouserve the picture, Kavitha can reach her school in two ways. From A, B roads which path will you suggest her to reach the school early? You will suggest the shortest distance pat*” which is road A.
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 3
Answer:
Road A

7th Class Science Textbook Page No. 72

Question 18.
What can we call the duration of motion of the object when it is moving from one place to another place?
Answer:
Time to travel

Question 19.
How can we call the period between starting and ending incidents (events) in a running race.
Answer:
Time.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 20.
How do we measure or estimate time?
Answer:
Clocks (watches) are the most common time measuring devices.

7th Class Science Textbook Page No. 73

Question 21.
Are all the motions same?
Answer:
No, all the motions are not same.

Question 22.
How many types of motions are there?
Answer:
Motion is of three types :

  1. Translatory motion
  2. Rotatory motion
  3. Oscillatory motion.

Question 23.
Have you played with a toy car? How does it move if you push it forward?
Answer:
It moves in translatory motion.

Question 24.
Do all the parts of bus like wheels, headlights, windows etc., move through the same distance from building ‘A’ to building ‘B’?
Answer:
Yes, all parts of the bus moves along the bus.

Question 25.
Is the direction of motion of bus along a straight line or a curved line?
Answer:
In a straight line.

7th Class Science Textbook Page No. 74

Question 26.
Can you give some more examples of motion in which all points of moving body move through same distance in the same direction?
Answer:
Examples:

  1. Coconut falling from a tree.
  2. Movement of lift.
  3. Car travelling on a straight road.
  4. Movement of water from the tap.

Question 27.
If all the parts of a body move along the curved path. What type of motion it is?
Answer:
Rotatory motion.

Question 28.
Observe the following diagrams.
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 7
1) Do you find any similarity in the above motions?
2) What is the path of motion of each part of the body that moves? ’
Answer:

  1. In the above situations, the parts of the objects move in a curved path with respect to a fixed centre or a fixed axis.
  2. They are in rotatory motion with circular path.

7th Class Science Textbook Page No. 75

Question 29.
Can objects have simultaneous Translatory and Rotatory motions?
Answer:
Yes. The cycle tyre moving in a straight line possesses both translatory and rotatory motions.

Question 30.
Have you played with a see-saw? How does it move?
Answer:

  1. See – saw moves in up and down direction with a fixed point.
  2. This type of motion is called oscillatory motion,

Question 31.
How does the needle of swing machine moves?
Answer:
The needle of a sewing machine moves in ‘to and fro’ motion or back and forth motion.

7th Class Science Textbook Page No. 76

Question 32.
Which similarity do you find in the moving parts of the above instruments?
Answer:
All are moving in ‘to and fro’ motion.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 33.
Is the direction of motion constant? Are they moving in the same path again and again?
Answer:

  1. The direction of motion of the objects changes alternatively along the same path about a fixed point. Hence the direction of motion of the object not constant.
  2. Yes, they are moving in the same path again and again.

Question 34.
Does all motions take same time to cover same distance?
Answer:
No.

7th Class Science Textbook Page No. 77

Question 35.
Do you find any relation between distance and time?
Answer:
Yes, distance divided by time give speed the objects.

Question 36.
How to calculate the average speed?
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 8

Question 37.
Is there any other method to express the motion of objects?
Answer:
Yes, we can express the motion of objects by graphs.

7th Class Science Textbook Page No. 78

Question 38.
Have you seen launching of rocket in T.V, how does it move?
Answer:

  1. Rockets move in translatory motion.
  2. The exhaust gases coming out of the engine’s nozzle at the high speed push the rocket forward.

Question 39.
Do you know which Indian organization is working for it?
Answer:
ISRO (Indian Space Research Organization) is working for it.

Question 40.
Where do these rockets are launched?
Answer:
Rockets are launched at Satish Dhawan Space Centre (SDSC – SHAR), Sriharikota, Andhra Pradesh.

7th Class Science Textbook Page No. 80

Question 41.
Have you noticed what happens if you let the air out of a inflated balloon?
Answer:
The air comes out of balloon in one direction and the balloon moves on the opposite direction with the same speed.

Question 42.
Have you observed the movement of Diwali rocket cracker?
Is there any similarity between movements of rocket cracker and real rocket?
Answer:

  1. Diwali rocket cracker takes flight, when it is fired.
  2. Motion of the actual rocket is similar to it.

Question 43.
Have you ever feel wonder of knowing weather report in advance?
Answer:
Yes, weather report forecasting in advance is possible because of artificial satellites.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Question 44.
How google maps give route maps between two places?
Answer:

  1. Google maps is a internet based application to show maps and roots between two places.
  2. They map our location with the pictures and data provided by artificial satellites.
  3. By using artificial intelligence, they integrate the information and give route maps between two places.

Activities and Projects

Question 1.
Use a tennis ball or cricket to make it translatory motion, rotatory motion, and oscillatory motion. Write a note on it and discuss with your friends.
Answer:

  1. I took a tennis ball and throwed it towards my friend to catch it. The motion showed . by ball is translatory motion.
  2. Spun the ball on my finger. It shows rotatory motion.
  3. Now it is rolled over the ground towards my friend. It shows both translatory and rotatory motions.
  4. Now tied this ball to a thread and moved forward and backward. This shows the oscillatory motion.

Question 2.
Make a water clock or a sand clock and use it to know time.
Answer:
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 2

  1. I took two, clear plastic bottles that are the same size and shape.
  2. Took the caps off, glued them together, and let the glue dry.
  3. I Punched a hole through the middle of the glued-together caps.
  4. Screwed the cap onto the first bottle, just like we normally would.
  5. Filled the second bottle with dry sand.
  6. Screwed the empty bottle onto the sand-filled bottle.
  7. Tested my sand timer by turning it upside down to know whether the sand is flowing smoothly from one bottle to the next or not.
  8. Marked the time on the bottles at various levels of sand by observing the time using stop watch.
  9. Wrapped some tape around the neck.
  10. My sand clock is ready to use.

Activities

Activity – 1

Question 1.
Classify the objects given in the figure based on motion and rest and what is your inference?
AP Board 7th Class Science Solutions Chapter 5 చలనం – కాలం 3

Objects at restObjects in motion

Answer:

Objects at restObjects in motion
BagButterfly
TableBird
HouseMerry – go-round

Inference:

  1. “An object is said to be in motion if it changes its position with respect to its surroundings.”
  2. “An object is said to be at rest if there is no change in its position with respect to its surroundings.”

Activity – 2

Question 2.
How do you show that force is required to change the object from rest into motion, to change its direction and even to change from motion to rest?
Answer:
AP Board 7th Class Science Solutions Chapter 5 చలనం – కాలం 4
Take a football and go to your school ground with your friends. Stand in a triangle shape as shown in the figure.Ask your friend to kick the ball in your direction. You pass the ball to another friend. Ask him to stop that ball. Now answer the following questions.
1) What did your friend apply to make the ball move?
Answer:
Force

2) What did you apply to change the direction of the ball?
Answer:
Force

3) What did your Mend apply to stop the moving ball?
Answer:
Force

You all applied muscular force with your legs to do so.

From the above observation we can say that force is required, to change the object from rest into motion, to change its direction and even to change from motion to rest.

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Activity – 3

Question 3.
Write any activities to explain the distance and displacement.
Answer:
Try to measure the distance between your bench to the place where MDM is been served in your school.

How much it is? (Student has to measure and answer on their own)

Have you ever used the google maps to know the distance and direction of your desti-nation?

Now, know the distance between your native place to district head quarter using google maps?

How much it is? _____ km. (Student has to measure and answer on their own)
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 3

Observe the picture, Kavitha can reach her school in two ways. From A, B roads which path will you suggest her to reach the school early? You will suggest the shortest distance path which is road A.

From the above activity and examples, we can say that when an object travels from one place to another, the length of the path can be measured and it gives the distance between the two places. When an object travels along the straight-line path from one place to another place, it will reach faster. The length of the straight-line path is the shortest distance between two places, called displacement.

Activity – 4

Question 4.
Conduct 100 meter running race with the help of your physical education teacher. Use stop watch from lab or mobile phone to measure time taken to complete the race by each student. Fill the table given below.
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 4
Answer:

Name of the studentTime taken to complete race
1. Satyam1 m 5 sec
2. Venkatesh53 sec
3. Rama Raju46 sec
4. Balakrishna1m 12 sec
5. Narendra58 sec
6. Vijay1 min

[Note : Fill in the table with your records)

AP Board 7th Class Science Solutions 5th Lesson Motion and Time

Activity – 5

Question 5.
Consider the following situation. Robert and Kamala measured the distance trav¬elled while coming to school and the time taken to cover that distance. Their data is shown in the Table. You can draw a graph by following the steps given below:

Time (Minutes)Distance (Meters)
1. 00
2. 1100
3. 2200
4. 3300
5. 4400
6. 5500

Answer:
1) Draw two perpendicular lines to represent the two axic anti mark them as OX and OY. Mark OX as time axis and OY as distance axis.

2) Choose a scale to represent distance and time on the graph.
Time : 1 min = 1 cm.
Distance : 50 m = 1 cn

3) Mark values for the time and the distance on the respective axis according to the scale you have chosen.

Now you have to mark the points on the graph paper to represent each set of values and, join al! the points on the graph. This is the distance-time graph for the given motion.

4) If the distance-time graph is a straight line, it indicates that the object is moving with a constant speed. However, if the speed of the object keeps changing, the graph can be of any other shape. From the distance-time graph we can find the speed of object.
AP Board 7th Class Science Solutions 5th Lesson Motion and Time 5

From the distance-time graph we can find the speed of object.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

SCERT AP 7th Class Science Study Material Pdf 4th Lesson Respiration and Circulation Textbook Questions and Answers.

AP State Syllabus 7th Class Science 4th Lesson Questions and Answers Respiration and Circulation

7th Class Science 4th Lesson Respiration and Circulation Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. Respiration is the process essential for ______ of living things.
2. Inhaled air has ____ oxygen and _____ has carbon dioxide.
3. ______ can breathe both with lungs and skin.
4.The latest respiratory disorder is
1. survival
2. more, less
3. Frog
4. COVID- 19

II. Choose the correct answer.

1. Colour of Haemoglobin is
a) Colourless
b) Blue
c) Red
d) Green
Answer:
c) Red

2. Lime water turns milky when it reacts with
a) Oxygen
b) Nitrogen
c) Carbon
d) Carbon dioxide
Answer:
d) Carbon dioxide

3. The part of respiratory system which branches into
a) Nasal cavity
b) Bronchi
c) Lungs
d) Trachea
Answer:
b) Bronchi

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

4. Humans normally breathe times per minute.
a) 14 to 20 times
b) 20 to 30 times
c) 72 times
d) Upto 80 times
Answer:
a) 14 to 20 times

III. Matching

A) Gills1. Earthworm
B) Tracheae2. Whale
C) Lungs3. Stem
D) Skin4. Fish
E) Stomata5. Cockroach
F) Lenticels6. Leaf
7. Flower

Answer:

A) Gills4. Fish
B) Tracheae5. Cockroach
C) Lungs2. Whale
D) Skin1. Earthworm
E) Stomata6. Leaf
F) Lenticels3. Stem

IV. Answer the following questions.

Question 1.
What is respiration?
Answer:
The oxygen reacts with the glucose present in digested food and breaks it down into Carbon dioxide and water to release energy. This process is called Respiration.
Glucose + Oxygen → Carbon dioxide + water + energy

Question 2.
Name the two types of respiration. Write its word equation.
Answer:
Respiration is primarily of two types (a) Aerobic (b) Anaerobic
(a) Aerobic respiration: oxygen involves in this process
Glucose + Oxygen → Carbon dioxide + water + Energy

(b) Anaerobic respiration: oxygen does not involve in this process
Glucose → Carbon dioxide + alcohol + Energy

Question 3.
Write the differences in composition of inhaled and exhaled air.
Answer:

CompositionQuantity in
inhaled air (in %)
Quantity
in exhaled air (in %)
Oxygen2115
Carbon dioxide0.04 44
Nitrogen7878
Water vapour0.963

Question 4.
Explain in detail the pathway of respiration in humans with the help of a flowchart.
Answer:
Parts that form the pathway of air in human respiratory system are
1) Nostrils,
2) Nasal cavity,
3) Pharynx,
4) Wind pipe,
5) Bronchi,
6) Lungs.
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 2

Question 5.
What are the different respiratory organs seen in animals and their method of working?
Answer:
The organs of respiration are different in different organisms.
A. Tracheae :

  1. Respiration that take place via tracheae is called Tracheal respiration.
  2. This is present in all insects. In this system there are openings called spiracles which enter the body of the insect through a network of air tubes called Tracheae.
  3. The tracheae reach all parts of the body and help in the exchange of gases.
  4. Ex : Grasshopper, Cockroach, Honey bee etc.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 4

B. Skin :
Cutaneous respiration :

  1. In some animals the skin is moist and slimy with mucus which helps in breathing through the skin. Ex : Earthworm.
  2. Frogs have lungs for breathing which they use when they are on land.
  3. But can breathe through the skin which is moist and slippery while in water.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 5

C. Gills:
Branchial respiration :

  1. Gills are seen in fishes. Gills are present beneath membranous covers on either side of the head.
  2. The gills have a rich supply of blood, for exchange of oxygen and carbon dioxide.
  3. Fish take in water through their mouth which passes out over the gills, the specialized organs to absorb the dissolved oxygen present in the water.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 6

D. Lungs:

  1. Respiration through lungs is called Pulmonary respiration.
  2. In all land animals and some water animals the organs for breathing are lungs.
  3. They are meant for taking oxygen from the air. Ex : Cow, dog, whale, humans etc.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 3

Question 6.
If the diaphragm and ribs do not expand and contract what will be the consequences?
Answer:

  1. A large thin muscular sheet called diaphragm is attached to the lower side of the ribcage and forms the floor of the chest cavity.
  2. The process of breathing involves the movement of the diaphragm and the ribcage, in men and women respectively.
  3. If they can’t contract inhalation and exhalation is not possible
  4. It means we are not able to breath and it Leads to death

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 7.
Write a report on the lab activity done in your school to prove that carbon dioxide is released during exhalation.
Answer:
Aim : To prove that carbon dioxide is released during exhalation What you need: Two small beakers .straw, lime water
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 7

How to do:

  1. Take two beakers. Label them as A and B.
  2. Fill two beakers up to half with clear lime water.
  3. Blow air into beaker ‘A’ using a straw..
  4. Pass atmospheric air repeatedly into beaker ‘B’ using a dropper.
  5. Observe the colour change in both the beakers.

What you see:
Lime water is clear and colourless in beaker B , but it turns milky white in beaker A when it reacts with carbon dioxide.

What you learn:
This concludes that there is more carbon dioxide in exhaled air when compared to inhaled air, it is proven that exhaled air contains carbon dioxide

8. Prepare slogans to create awareness about a) evil effects of smoking b) COVID-19 prevention.
Answer:
a) Evil effects of smoking :
Smoking is injurious to health.
Say YES to NO SMOKING
Be smart, Don’t start
Smoking? You must be joking
You don’t have to smoke to be hot!
Be Cool – Don’t Be a Smoking Fool
Don’t take smoke as a joke
Smokers are jokers
Don’t put your lips on fire

b) COVID-19 prevention
Follow the SMS always
SMS means social distance, mask
and sanitization
Don’t neglect be cautious
Wash the hand regularly
Break the chain to kill the corona
Minimum distance is maximum safe

Question 9.
Make use of the Stethoscope made by you to measure the number of heart beats of five of your friends and note down your findings in the table given.
Answer:
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 8

7th Class Science 4th Lesson Respiration and Circulation InText Questions and Answers

7th Class Science Textbook Page No. 53

Question 1.
How fish could breathe in water?
Answer:
Fish has gills, the specialized organs to absorb the dissolved oxygen present in the water. Hence, fish breathe through gills.

Question 2.
Why humans and other animals cannot breathe in water?
Answer:

  1. Humans, and some other animals have only lungs for breathing.
  2. But lungs cannot absorb the oxygen from the water.
  3. ence, they cannot take breathe in water.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 3.
What is breathing?
Answer:
The process of inhalation and exhalation of air is called Breathing.

Question 4.
Inhale deeply. Do you feel air moving inwards?
Answer:
Yes.

Question 5.
Now hold a finger under your nose and slowly release the air. How do you feel now? Do you feel air moving outwards?
Answer:
Yes.

Question 6.
How does the air reach the lungs? What are the organs involved in this process?
Answer:
a) Air reaches the lungs from nostril through various organs.
b) The organs involved in this process are 1) nostril 2) nasal cavity 3) pharynx 4) wind pipe 5) bronchi 6) lungs.

7th Class Science Textbook Page No. 54

Question 7.
How are the lungs expanding and contracting?
Answer:

  1. When we breathe in, the chest along with ribcage moves upwards and air enters into our lungs. So, lungs expand in inspiration.
  2. When we breatheout, the chest along with ribcage moves downwards and air goes out. So, lungs contracts in expiration.

7th Class Science Textbook Page No. 56

Question 8.
Why does the amount of oxygen vary between inhaled and exhaled air?
Answer:
Because the oxygen in inhaled air is absorbed in the lungs.

Question 9.
Which gas has increased in quantity in the exhaled air? Why?
Answer:

  1. The quantity of carbondioxide is more in the exhaled air.
  2. Because carbondioxide from the blood is transferred to the exhaled air.

Question 10.
In which beaker does the lime, water turned milky white?
Answer:
Beaker ‘A’.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 11.
What does this change indicate?
Answer:

  1. Carbondioxide turns lime water milky.
  2. The change indicates that beaker ‘A’ contains more carbondioxide.
  3. This concludes that there is more carbondioxide in exhaled air when compared to inhaled air.

Question 12.
What happens to the air in the lungs?
Answer:
Oxygen from the inhaled air is absorbed by the blood vessels present in the lungs. Carbon dioxide collected by the blood vessels from all parts of the body enters the Lungs. The absorbed oxygen is transported to every part (Cell) of our body.

7th Class Science Textbook Page No. 57

Question 13.
Checklist.
Tick (✓) the statements which you think are correct and (✗) for wrong me.
1) There is no harm in trying a cigarette once, because one can stop after that. [ ✗ ]
2) Smoking a cigarette once a day is not harmful. [ ✗ ]
3) One can stop smoking only with his will power. [ ✓ ]
4) Smoking helps you feel good and relaxed. [ ✗]
5) Smoking is not harmful to health. [ ✗ ]

Question 14.
What about the smoke inhaled by smokers?
Answer:

  1. The smoke inhaled by smokers is being sent to all parts of the body.
  2. Tobacco smoke contains a highly dangerous substance called Nicotine.
  3. This poisonous substance is also carried to all cells of the body. . ‘
  4. Smoking leads to lung cancer, tuberculosis and other respiratory disorders.

Question 15.
Do all animals have the same type of respiratory organs?
Answer:
No.

Question 16.
What are the respiratory organs in whales?
Answer:

  1. Whales breathe air into their lungs,
  2. They cannot breathe under water like fish can as they do not have gills.
  3. They breathe through nostrils called a blowhole, located right on top of their heads.

Question 17.
Why is the skin of frog always wet and slimy?
Answer:

  1. In some animals like frog the skin is moist and slimy with mucus.
  2. It helps in breathing through the skin while in water.

7th Class Science Textbook Page No. 58

Question 18.
How do plants breathe?
Answer:

  1. Plants take in oxygen and leave out carbondioxide in breathing.
  2. They breathe through stomata, lenticels and root hairs.

Question 19.
What are their (plants) respiratory organs?
Answer:
The respiratory organs in plants are

  1. Stomata in leaves
  2. Lenticels in stems
  3. Root hairs in roots.

7th Class Science Textbook PageNo. 59

Question 20.
What changes did you observe in the lime water?
Answer:
The lime water turns milky white.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 21.
Why was there a change in the lime water?
Answer:
The carbondioxide, which was released by the plants (sprouts)’is turned lime water milky white.

Question 22.
How does blood reach al! parts of the body?
Answer:

  1. Our circulatory system consists of heart, blood vessels and blood.
  2. There are three types of blood vessels – arteries, veins and blood capillaries.
  3. These blood vessels are connected to the heart.
  4. While heart is pumping the blood, the blood capillaries which are connected to the arteries with the veins distribute the blood to the body parts.

Question 23.
How is the oxygen and glucose absorbed by the blood transported to all the parts of our body?
Answer:

  1. Oxygen and glucose absorbed by the blood transported to all parts of the body through circulatory system.
  2. The circulatory system consists of heart, blood vessels and blood.
  3. Blood absorbs the glucose (Digested food) from the digestive system, and oxygen from lungs.
  4. This blood is pumped by heart to all parts of the body through a kind of blood vessels called arteries.
  5. These arteries ends in another type of blood vessels called blood capillaries.
  6. These capillaries are thin arid narrow.
  7. So, through their thin walls, blood is distributed to all body parts.
  8. Now oxygen and glucose absorbed by the blood transported to the body parts.

7th Class Science Textbook Page No. 61

Question 24.
Is the blood in all organisms same as In humans?
Answer:
No, the blood in all organisms is not same as in humans.
For example, the colour of human blood is red whereas blue in prawns.

7th Class Science Textbook Page No. 62

Question 25.
Which new disease created a global impact recently?
Answer:
COVID -19 was the new disease that created a global impact recently.

Question 26.
What is a Pandemic?
Answer:
A disease which infects most of the people in a country or the whole world at the same time is called a Pandemic.

Question 27.
How can we prevent it?
Answer:
We can prevent any disease or infection by improving our immunity.

7th Class Science Textbook Page No. 63

Question 28.
Why do we sneeze?
Answer:

  1. Sneezing occurs when we inhale air with dust, smoke, pollen or strong smells.
  2. Sneezing expels the unwanted, harmful substances from the lungs.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 29.
Why do we yawn?
Answer:

  1. Yawning is caused when the respiratory rate gets slowed down resulting in insufficient supply of oxygen to the brain.
  2. To overcome this situation, the body goes for the involuntary opening of the mouth to take in a long deep breathe of air.

7th Class Science Textbook Page No. 64

Question 30.
What is first aid?
Answer:
The simple assistance provided to a person suffering from injury or serious situations before a doctor attending is called First Aid.

Question 31.
What is the first aid tor drowning?
Answer:
When a person drowns we should first bring him out of the water and make him lay down on the back with the face turned to one side. Check for any sand or mud in the nose, mouth and ears and remove them.

Now, slowly press the abdomen to create pressure on the diaphragm and on the lungs to push out the water that has entered the lungs. Repeat the process by turning the person to lie on the abdomen until recovery. Give some warm clothing and hot tea on recovery.

Think & Respond

Page No. 54

Question 1.
Why is the right lung larger than the left lung?
Answer:

  1. Heart is located in the centre of the chest cavity slightly bent towards the left.
  2. So, as a result the left lung being smaller than the right lung.

Activities and Projects

Question 1.
Make use of a water bottle with water and two straws to test your lungs power.
Answer:

  1. Take a bottle with a lid.
  2. Make two holes in the lid and insert two straws through them.
    AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 9
  3. Fill the bottle with water and close its mouth with the lid along with straws.
  4. Adjust the straws such that one should be till the bottom of the bottle and other till the surface of the water.
  5. Now blow air with force through the straw placed above the water surface of water.
  6. As this blowing air put pressure on water, it escapes out through the second straw.
  7. The hight of the water raises through the second straw or the empty space formed in the bottle indicates your lung power.

Question 2.
Make use of a mirror to know whether our breath contains water vapour as given in the table 2 showing what our breath contains.
Answer:

  1. Take a mirror in to your hands during morning and blow air on to that.
  2. You can find that your image in the mirror become blarred.
  3. Now wipe the mirror with your hand.
  4. You will find that the image become clear and some moisture touches to your hand.
  5. This moisture is nothing but the water vapour present in your breath.
  6. This phenomena can be observed more clearly during winter mornings than summer mornings.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Question 3.
Prepare a model of lungs using a water bottle, balloons and Y tube to show the importance of diaphragm in respiration.
Answer:
Materials:
Y-shaped tube, a large balloon, two small balloons, a one litre plastic bottle, cork.
AP Board 7th Class Science Solutions Chapter 4 శ్వాసక్రియ – ప్రసరణ 10

Method of construction :
Cut the plastic bottle to half its size. Fix two small balloons at both ends of the Y-tube. Make a small hole in the cork and fix the Y-tube through the hole as shown in the picture. Cut the large balloon into two halves and fix one half tightly around the open part of the bottle.

Method of working :
Hold the balloon at the middle and pull it slowly downwards. The balloons become inflated (Expands), Now leave the balloon free. Then the balloons collapse (contracts).

The expansion and contraction of the lungs is almost like this. Similar to the balloon tied firmly at the base of the bottle, there is a part called ‘diaphragm’ in our body to control the expansion and contraction of the lungs.

Activities

Activity – 1

Question 1.
How do you perform breathing by an activity?
(OR)
Do an activity to show the stages in breathing?
Answer:

  1. Inhale deeply.
  2. You may feel air moving inwards.
  3. The process of inhaling air is called inspiration.
  4. Now hold a finger under your nose and slowly release the air.
  5. You may feel air coming outside.
  6. The process of exhalation of air is called expiration.
  7. This activity shows us that there are two stages in breathing – Inhalation and Exhalation.

Activity – 2

Question 2.
How do you prove that there is a difference in the size of the chest cavity during inhalation and exhalation process?
(OR)
Fill in the table with an activity.
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 10
Answer:

1. Take a measuring tape.
2. Hold the tape around the chest of one of your friend and measure the width of her/ his chest.
3. Hold the tape lightly and ask your friend to breathe in and again measure the width of her/his chest.
4. Repeat it with more friends and record your observations in the given table.
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 11
5. We can conclude from this activity that there is a difference in the size of the chest cavity during inhalation and exhalation process.

Activity – 3

Question 3.
How do you prove that there is more carbon dioxide in exhaled air when compared to inhaled air?
Answer:
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 7

  1. Take two beakers. Label them as A and B.
  2. Fill two beakers upto half with clear lime water.
  3. Blow air into beaker ‘A’ using a straw.
  4. Pass atmospheric air repeatedly into beaker B’ using a dropper.
  5. Observe the colour change in both the beakers.
  6. Lime water is clear and colourless, but it turns milky white when it reacts with carbon dioxide.
  7. This concludes that there is more carbon dioxide in exhaled air when compared to inhaled air.

Activity – 4

Question 4.
How do you prove that the sprouting seeds respire to take in oxygen and leave out carbon dioxide?
Answer:
Take a wide mouthed bottle and place a handful of sprouting seeds in it. Prepare some fresh lime water in a small container and place it carefully in one corner of the bottle. Close the cap of the bottle and apply vaseline on the edges to make it air-tight. Leave the apparatus undisturbed for a day or two. After two days open the cap and carefully take out the lime water container and observe the changes.
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 12

1. What changes did you observe in the lime water?
Answer:
The lime water changes into milky white colour.

2. Why was there a change in the lime water?
Answer:
The sprouting seeds respire to take in oxygen and -leave out carbon dioxide. The lime water in the small container reacts with the carbon dioxide released by sprouting seeds to change into milky white colour.

AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation

Activity – 5

Question 5.
How do you make a stethoscope? What are the material do you require?
(OR)
How do you know the rate of heart beat?
Answer:
Aim : To know the rate of heart beat.

What we need :
Rubber tube, Y – shaped attachment, small funnel, rubber sheet /balloon, steel tongue cleaner, beads or earphone buds, insulation tape.

Procedure:
1. Insert rubber tubes to the three ends of a Y shaped attachment.
2. Arrange a small funnel to the rubber tube attached to the lower single arm of the Y shaped attachment.
3. Tie a rubber sheet / balloon to the broad end of the funnel.
AP Board 7th Class Science Solutions 4th Lesson Respiration and Circulation 13
4. Arrange ear buds to the ends of rubber tubes attached to the upper two arms of the Y shaped attachment.
5. Arrange a steel tongue cleaner connecting the two upper tubes over the Y shaped attachment to give support.

Working:
Keep upper two rubber tube ends in your ears. Put the funnel on the chest of your friend and observe the sound of heart beat.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

SCERT AP 7th Class Science Study Material Pdf 3rd Lesson Nutrition in Organisms Textbook Questions and Answers.

AP State Syllabus 7th Class Science 3rd Lesson Questions and Answers Nutrition in Organisms

7th Class Science 3rd Lesson Nutrition in Organisms Textbook Questions and Answers

Improve Your Learning

I. Fill in the blanks.

1. During photosynthesis ________ gas is released.
2. The tiny pores present on the surface of the leaves are __________
3. ____________ is the outermost layer of teeth.
Answer:
1. oxygen
2. stomata
3. Enamel

II. Choose the correct answer.

1. Muscular tubular structure that connects pharynx with Stomach
a) Duodenum
b) Buccal cavity
c) Oesophagus
d) Rectum
Answer:
c) Oesophagus

2. Non insectivorous plant,
a) Drosera
b) Nepenthes
c) Utricularia
d) Dodder
Answer:
d) Dodder

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

3. The green colour pigment in the leaf
a) chloroplast
b) stomata
c) chlorophyll
d) all the above
Answer:
c) chlorophyll

III. Matching

A) Autotrophs1. Mushroom
B) Saprophytes2. Mango plant
C) Parasite3. Food vacuole
D) Holozoic4. Intestinal worms
E) Amoeba5. Man
6. Egestion

Answer:

A) Autotrophs2. Mango plant
B) Saprophytes1. Mushroom
C) Parasite4. Intestinal worms
D) Holozoic5. Man
E) Amoeba3. Food vacuole

IV. Answer the following questions.

Question 1.
Distinguish between autotrophic and heterotrophic nutrition.
Answer:

Autotrophic nutritionHeterotrophic nutrition
1. The mode of nutrition in which organisms make food by them selves is called autotrophic nutrition1. The mode of nutrition in which organisms depend on other organisms for food is called Heterotrophic nutrition
2. Organisms that perform autotro phic nutrition are called autotrophs.2. Organisms that perform heterotrophic nutrition are called heterotrophs.
3. They need carbon dioxide, water and sunlight to prepare their own food.3. They don’t need these materials as they feed on other organisms for food.
4. They are green in colour due to the presence of a colouring pigment called chlorophyll.4. They don’t have chlorophyll.
5. Ex: Plants5. Ex: Animals

Question 2.
What is Photosynthesis? Write its word equation.
Answer:
The process by which green plants make their own food from carbon dioxide and water by using light energy in the presence of chlorophyll is called Photosynthesis.
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 1a

Question 3.
Describe different types of teeth and their functions.
Answer:
There are four different types of teeth in our human beings.
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 2
1. Incisors :
The four front teeth in both the upper and lower jaws are called incisors. Thus there are eight incisors in the oral cavity. They have sharp incisal edge. Their primary function is to cut food.

2. Canines :
They are behind and adjacent to the lateral incisors on both the jaws. There are four canines in the oral cavity. They have single sharpcusp. Their main function is to tear food.

3. Premolars :
These teeth are located behind and adjacent to the canines. There are two premolars on the upper jaw and also two premolars on the lower jaw on either side of the mouth. Thus there are eight premolars in the oral cavity.These teeth can have 3-4 cusps. They are designed to crush food.

4. Molars :
The most posterior teeth in the mouth are the molars. There are 12 molars in the permanent dentition.They have broader and flatter surfaces with 4-5 cusps. They are designed to grind food.

Question 4.
What happens if leaves of a green plant are coated with green paint?
Answer:

  1. If the leaves of a green plant are coated with green paint, then stomata present on the surface of the leaf will get blocked.
  2. As a result exchange of gases will be affected.
  3. So, plant may not get sufficient amount of carbon dioxide for photosynthesis and oxygen for respiration.
  4. oreover, paint over the leaf prevents the reaching of sunlight to the leaf.
  5. As a result plant will not be able to perform photosynthesis and respiration, gradually it leads to its death.

Question 5.
What questions will you ask a doctor to know about gastritis?
Answer:

  1. What is gastritis?
  2. What are the reasons for gastritis?
  3. How can it be prevented?
  4. Is there any relation between gastritis and life style?
  5. Are there any home remedies to get relief from gastritis?
  6. What is the treatment for gastritis?

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 6.
How can you prove that the leaves other than green also carry out photo synthesis? (Activity – 2)
Answer:
Aim: To know whether leaves other than green perform photosynthesis or not.

What you need:
Red or brown coloured leaves, dropper; test tube, iodine solution, water.

What to do:
Take few Red or brown coloured leaves, add few drops of water and mash them to make a paste, collect the extract by squeezing the paste (pulp). Take 5-6 drops of this extract into a test tube. Also add two drops of iodine solution. Note your observations.

What do you see:
The colour of the leaf extract turns blue black.

What do you learn:
This shows the presence of starch in the leaves, confirming that red or brown coloured leaves also perform photosynthesis.

These non-green leaves also have chlorophyll. The large amount of red, brown and other pigments mask the green colour. So photosynthesis takes place in these leaves also.

Question 7.
Draw a neat labelled diagram of human digestive system.
Answer:
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 3

Question 8.
Draw a flow chart showing nutrition in amoeba.
Answer:
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 4

Question 9.
How will you appreciate the role of saprophytes in cleaning of the earth surface?
Answer:

  1. Saprophytes grow on dead and decaying matter.
  2. They secrete digestive juices on it, convert it into a solution and then absorb the nutrients from it.
  3. During this process they decompose them and mix them with the soil.
  4. In this Way they help in cleaning the earth surface by removing the dead and decaying matter.
  5. This process help in recycling the nutrients too.
  6. Thus they are doing a great service to the entire living world.

Question 10.
What precautions will you take to keep your digestive system healthy?
Answer:
To keep my digestive system healthy, I will

  1. take simple and balanced diet
  2. drink sufficient quantity of water.
  3. brush my teeth regularly.
  4. prefer natural foods.
  5. avoid Junk foods and cool drinks.
  6. never smoke.
  7. never consume alcohol.
  8. never chew tobacco and tobacco products.
  9. do moderate exercises regularly.
  10. take fibre rich food.

7th Class Science 3rd Lesson Nutrition in Organisms InText Questions and Answers

7th Class Science Textbook Page No. 37

Question 1.
My mother prepares food for me. Who prepares food for animals?
Answer:

  1. Animals cannot make food.
  2. They depend on other plants and animals for food.

Question 2.
How do animals get their food?
Answer:
Animals get their food from plants and animals.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 3.
Plant is a living thing. What is the food for it?
Answer:
Plants also need food, such as glucose, carbohydrates, minerals etc.

Question 4.
How do they get their food?
Answer:

  1. Plants prepare their food by using photosynthesis.
  2. Some plants get their food from other plants and animals.

Question 5.
What is the food for mushrooms?
Answer:
Mushrooms get their food from dead and decaying matter.

Question 6.
Even you might have got the same doubts?
Answer:
Yes.

Question 7.
Is the nutrition in plants same as in animals?
Answer:
No.

Question 8.
We eat food daily. But plants are not eating food then how are they living?
Answer:
Plants are able to live as they make their food themselves.

7th Class Science Textbook Page No. 38

Question 9.
How can the green plants prepare their own food?
Answer:
Green plants make their own food from carbon dioxide and water by using light energy in the presence of chlorophyll. This process is called photosynthesis.

7th Class Science Textbook Page No. 39

Question 10.
In which part of the plant, does photosynthesis takes place?
Answer:
The synthesis of food occurs in all green parts of plant body. For Ex : leaves.

Question 11.
How do the raw materials required for photosynthesis reach there?
Answer:

  1. Plants get carbon dioxide from air through stomata in leaves.
  2. Water obsorbed by the roots are transported to the leaves through the stem.
  3. Chlorophyll captures the energy of the sunlight.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 12.
Do they perform photosynthesis?
Answer:
Yes, the plants with red and brown coloured leaves make their food in the process of photosynthesis.

7th Class Science Textbook Page No. 40

Question 13.
What is the importance of sunlight in photosynthesis?
Answer:

  1. Sunlight is the source of energy in the process of photosynthesis.
  2. The solar energy is captured by the leaves and stored in the plant in the form of food.
  3. If there is no sunlight starch cannot be prepared in plants.

Question 14.
Plants get water from the soil through their roots. How does this water reach the leaves from the roots? What path does it follow?
Answer:

  1. Water absorbed by the roots are transported to the leaves through the stem.
  2. It follows the path from roots to leaves via stem.

7th Class Science Textbook Page No. 41

Question 15.
How do carbondioxide enters the leaves?
Answer:
Carbondioxide from air is entered the leaves through stomata which are present in the surface of the leaf.

Question 16.
How do oxygen comes out of the leaves?
Answer:
Through the stomata, oxygen produced in photosynthesis goes out of the leaves.

Question 17.
Where do these substances (starch, fats and proteins) come from?
Answer:
Plants produce sugar first, which is converted to starch and then other compounds like fats and proteins.

7th Class Science Textbook Page No. 42

Question 18.
There are some plants which do not have chlorophyll. How do they get their food?
Answer:

  1. Organisms which do not have chlorophyll, cannot synthesis their food. They are heterotrophs.
  2. Heterotrophic nutrition is of different types basing on how they obtain their food from other organisms.
  3. Some plants depends on dead and decaying matter and some on other plants.

Question 19.
How do they (mushrooms) get their food?
Answer:

  1. Mushrooms grow on dead and decaying matter.
  2. They secrete digestive juices on it convert it into a solution and then absorb the nutrients from it.
  3. This type of nutrition is called saprophytic nutrition.

Question 20.
A rat died at Pullaiah’s house. What happens to the body after few days? Do you find any traces of its body after a month? Where did that body go?
Answer:

  1. Saprotrophs like fungi and bacteria will grow on the dead body of the rat, after a few days.
  2. We cannot find any traces (except strong bones) of the dead body of rat, after a month.
  3. The decomposed body of the rat is mixed with the soil.

Question 21.
How does Cuscuta survive? From where does it get nutrition?
Answer:

  1. Cuscuta take readymade food from the plant on which it is climbing.
  2. It develops special roots called haustoria, which penetrate into the tissues of the host plant and absorb food materials from them.

7th Class Science Textbook Page No. 43

Question 22.
How do animals take their food?
Answer:

  1. Animals obtain their food from other organisms.
    Ex : Cow eats grass, cat drinks milk, dog eats meat.
  2. They take the food in the form of solid or liquid.

Question 23.
Where do they digest it?
Answer:

  1. Animals take food into the body for digestion.
  2. Digestion occurs inside the body.

7th Class Science Textbook Page No. 44

Question 24.
What do you understand from Sabiha observations?
Answer:
Holozoic nutrition is the mode of heterotrophic nutrition in which the food is taken in solid or liquid form from the outside and is digested inside the body.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 25.
How do Amoeba takes its food?
Answer:

  1. Amoeba has many small bubbles – like vacuoles in its cytoplasm.
  2. Amoeba pushes out one or more finger – like projections called pseudopodia for movement and capture of food.
  3. Food vacuole forms around the captured food.
  4. Food get digested in it, absorbed into the cytoplasm and assimilates.
  5. Finally, undigested food is sent out by opening this vacuole out at the body surface.

Question 26.
Is the nutrition in human beings also holozoic?
Answer:
Yes. The nutrition in human beings also holozoic.

7th Class Science Textbook Page No. 45

Question 27.
Have you ever wondered what happens to the food inside the body?
Answer:
Seeing the food that went into the body being digested, I was very surprised.

Question 28.
Do you have any structures like vacuole in our body?
Answer:
No.

Question 29.
Where do the food eaten by us go?
Answer:

  1. We take food in the form of solids or liquids.
  2. It goes to the digestive system, which is consisting of the alimentary canal and digestive glands.
  3. Food get’s digested in the digestive system.
  4. Digested food is absorbed by the blood.
  5. Blood transports digestive food to different parts of the body for assimilation.
  6. Undigested food will be ejected out of the body.

Question 30.
What change does the food undergo inside the mouth?
Answer:

  1. We chew the food with the teeth and break it down mechanically into small pieces.
  2. Carbohydrates are digested in the mouth.
  3. Food also mixed with saliva and thrown into the stomach.

7th Class Science Textbook Page No. 46

Question 31.
How many kinds of teeth could you find?
Answer:
Four types of teeth.

Question 32.
Which teeth do you use for biting and cutting and which one for tearing?
Answer:

  1. Teeth which are located at front side of the mouth are used for biting and cutting. (These are called incisors.)
  2. Canines are used for tearing, which are located behind incisors.

Question 33.
Have you noticed formation of new teeth in the mouth of small children? Do all the teeth, wfiich sprouted first, remain forever?
Answer:

  1. Yes. f have noticed that the formation of new teeth in the mouth of small children.
  2. The teeth which sprouted first do not remain forever.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 34.
Do you know how does tooth decay happens?
Answer:
When food remains stuck in teeth, bacteria feed on them. As a result, lactic acid is produced, which causes the destruction of enamel and it leaves toothache.

7th Class Science Textbook Page No. 47

Question 35.
How does acid come into contact with teeth?
Answer:
When food remains stuck in teeth, bacteria feed on them. Bacteria produce lactic acid. This acid come to contact with teeth.

Question 36.
Do you now realise the importance of cleaning your teeth after food?
Answer:
Yes. It is very important. If we do not clean odr teeth properly after food, some food material will stuck in between teeth and causes tooth decay.

Think & Respond

7th Class Science Textbook Page No. 39

Question 1.
Testing tiie starch in the leaf directly with iodine had certain problems. To obtain results successfully think and discuss regarding this with your teacher.
Answer:

  1. Leaves are green in colour.
  2. When iodine solution is put on a leaf, it should turn blue if starch is present.
  3. However, the green colour of the leaf disguises the blue colour.
  4. More over cuticle present on the leaf hardly allows the iodine to reach he starch present in side the leaf.

7th Class Science Textbook Page No. 47

Question 2.
Which habit should be practised for the health of teeth? Why?
Answer:

  1. Brush teeth twice a day with fluoride tooth paste.
  2. Use dental floss to clean between your teeth.
  3. Change your tooth brush every 3 months.
  4. Cut down on how often you have sugary foods like chocolates, sweets and cool drinks. Visit dentist regularly.

The above habits should practise for healthy teeth. Because, if we do not clean our teeth and mouth after eating, many harmful bacteria begin to live and grow in mouth. These bacteria breakdown the sugars present from left over food and release acids. These acids gradually damage teeth and causes severe toothache.

Activities and Projects

Question 1.
Designer leaves making select any broad-leaved potte d plant. Cut a card board with a design of your choice and seal the selected leaf with the card board. Let the plant stand under the sun for a week then remove the card board. You will get designer leaves plant. Try to make more leaves with designs and display your plant but don’t forget to present your writeup.
Answer:
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 5
I have selected a potted plant with broad leaves.

I cut a design of my name first letter on a card board.

I sealed the selected leaf with the card board.

I kept this plant under the sun for a week.

Then I remove card board.

I surprised to see the design of my name first letter on the leaf.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Question 2.
Collect small plants, or the branches or other parts of big plants from your locality and classify them into autotrophs, parasites, saprophytes and symbionts. With the help of your teacher, preserve them in the form of specimens in the biology laboratory of your school.
Answer:
I have collected the following plants and classified them in to following categories.

Autotrophs:
Neem, mango, hibiscus, Paddy

Parasites:
Cuscuta, loranthus.

Saprophytes:
Mush rooms, Moulds Symbionts: Red gram, beans,

I preserved these specimens in the biology laboratory of our school.

Activities

Activity – 1

Question 1.
Fill the following table basing on your own observations and information collected from elders.
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 6

Answer:

Name of the organismPrepare own food/ Depend on other organisms for foodInference (Autotroph/ Heterotroph)
1. Mango treePrepares own foodAutotroph
2. CatDepends on other organismsHeterotroph
3. Rose plantPrepares own foodAutotroph
4. MushroomDepends on other organismsHeterotroph
5. LeechDepends on other organismsHeterotroph
6. GoatDepends on other organismsHeterotroph
7. Human beingsDepends on other organismsHeterotroph

→ Are all plants Autotrophs?
Answer:
No.

→ What kind of nutrition is seen in mushrooms?
Answer:
Heterotrophic nutrition (Saphrophytic).

→ What kind of nutrition is seen in animals?
Answer:
Heterotrophic (Parasitic, Holozoic).

7th Class Science Textbook Page No : 38 (Autotrophic Nutrition – Photosynthesis)

Question 2.
Anu said plants also prepare their food same as we prepare boiled rice.
Comparative table prepared by her was given below. Study the table and answer the questions.

Preparation of Boiled RicePreparation of food by green plants
Raw materialsRice, WaterCarbon dioxide, water
Source of energyFire from stoveSunlight
Happens inVessel/cookerChloroplast in green parts
Finally formsBoiled riceGlucose/Carbohydrates

→ What are the raw materials required for the preparation of food by green plants?
Answer:
Carbondioxide, water.

→ Which gas is taken by plants during this process?
Answer:
Carbondioxide.

→ Name the food material formed in plants?
Answer:
Glucose / Carbohydrates.

→ What do you conclude from the above information?
Answer:
Plants prepare their own food.

Activity – 3

Question 3.
How do you prove that the sunlight is essential for photosynthesis?
Answer:
Aim: To know the importance of light in photosynthesis.

What you need :
Two potted plants, dropper, test tube, iodine solution, water.

What to do:
Take two potted plants of the same kind. Keep one in the dark (or in a black box) for 72 hours and the other in the sunlight. Perform iodine test with the Leaf extracts of both the plants as you did in activity-2. Note your observations.

What do you see :
The colour of the leaf extract of first plant does not change. The colour of the leaf extract of second plant turns blue black.

What do you learn :
This shows the presence of starch in the leaves of plant kept in the sunlight indicating the occurrence of photosynthesis. There is no starch in the leaves of the plant kept in the dark. This confirms that sunlight is essential for photosynthesis.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Activity – 4

Question 4.
How do you conduct an activity to show the saprotrophic nutrition in bread moulds?
Answer:
Aim:
To observe the saprotrophic nutrition ih bread mould.

What you need:
Piece of bread, water, a container and a hand lens.

What to do:
Take a piece of Bread, in a container. Sprinkle some water and close the container. Open the container after few days and observe. (Use mask and gloves while doing this activity).

What do you see:
You will see cotton-like threads spread on the piece of bread and bread size is decreased.

What do you learn:
These thread-like structures are some sort of plants called Fungi. They don’t have chlorophyll so they obtain their food from dead and decaying matter.

Activity – 5

Question 5.
Wash your hands. Look into the mirror and count your teeth. Use your index finger to feel the teeth. How many kinds of teeth could you find? Take a piece of an apple, sugar cane or bread and eat it. Which teeth do you use for biting and cutting and which one for tearing?
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 2

Also find out the ones that are used for chewing and grinding? Compare your observations with the given picture and fill the table.
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 7

(Note: The picture of teeth on lower jaw was given here. The number and type of teeth on the upper jaw is same as on the lower jaw.)
AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms 8

→ What do you find?
Answer:
I found different types of teeth in my mouth.

→ How many teeth do you have? Is it equal to the number given in the picture?
Answer:
I have 28 teeth. No, in the figure there are 32 teeth. But 1 have only 28 teeth.

→ Which teeth are absent?
Answer:
I have no (wisdom teeth). 4 molars are absent in my mouth.

AP Board 7th Class Science Solutions 3rd Lesson Nutrition in Organisms

Activity – 6

Question 6.
How does acid damage our teeth? Prove with an activity.
Answer:
Aim : To know the process of tooth decay.

What you need :
Marbles, dilute hydrochloric acid and test tube.

What to do:
Put a few small pieces of marble in dilute hydrochloric acid. Examine after awhile.

What do you see :
Acid reacts with marble and dissolves it.

What do you learn :
The enamel, which is a calcium compound, reacts with acid and gets destroyed in the same way a marble reacts with hydrochloric acid.

When food remains stuck in teeth, bacteria feed on them. As a result, lactic acid is produced, which causes the destruction of enamel.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

Students get through AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు which are most likely to be asked in the exam.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

సాధించిన సమస్యలు
(Solved Problems)

ప్రశ్న 1.
కింది విలువలు కనుక్కోండి.
(i) sin \(\frac{5 \pi}{3}\)
(ii) tan (855°)
(iii) sec \(\left(\frac{13 \pi}{3}\right)\)
సాధన:
(i) sin \(\frac{5 \pi}{3}\) = sin (2π – \(\frac{\pi}{3}\))
= -sin\(\frac{\pi}{3}\)
= –\(\frac{\sqrt{3}}{2}\)

(ii) tan (855°) = tan (2 × 360° + 135°)
= tan (135°)
= tan (180° – 45°)
= -tan 45° = -1

(iii) sec \(\left(\frac{13 \pi}{3}\right)\)
= sec (4л + \(\frac{\pi}{3}\))
= sec \(\frac{\pi}{3}\)
= 2

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 2.
కింది వాటిని సూక్ష్మీకరించండి.
i) cot \(\left(\theta-\frac{13 \pi}{2}\right)\)
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 1

ii) tan \(\left(-23 \frac{\pi}{3}\right)\)
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 2

ప్రశ్న 3.
sin2 \(\frac{\pi}{10}\) + sin2 \(4\frac{\pi}{10}\) + sin2 \(\frac{6\pi}{10}\) + sin2 \(\frac{9\pi}{10}\) విలువను కనుక్కోండి.
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 3
= sin2 \(\frac{\pi}{10}\) + cos2 \(\frac{\pi}{10}\) + cos2 \(\frac{\pi}{10}\) + sin2 \(\frac{\pi}{10}\)
= 2 (∵ sin2 θ + cos2 θ = 1 ∀ θ)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 4.
sin θ = \(\frac{4}{5}\), θ మొదటి పాదంలో లేకపోతే cos θ విలువను కనుక్కోండి.
సాధన:
∵ sin θ = \(\frac{4}{5}\), θ మొదటి పాదంలో లేదు.
⇒ θ, రెండవ పాదంలో ఉంటుంది. . sin θ +ve కనుక,
⇒ cos θ = – \(\sqrt{1-\sin ^2 \theta}\) = –\(\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{-3}{5}\)
cos θ = –\(\frac{3}{5}\)

ప్రశ్న 5.
sec θ + tan θ = \(\frac{2}{3}\) అయితే, sin θ విలువ కనుక్కొని, θ ఏ పాదంలో ఉంటుందో కనుక్కోండి.
సాధన:
∵ sec θ + tan θ = \(\frac{2}{3}\)
⇒ sec θ – tan θ = \(\frac{3}{2}\) ; (∵ sec2 θ – tan2 θ = 1)
⇒ 2 sec θ = \(\frac{2}{3}\) + \(\frac{3}{2}\) = \(\frac{4+9}{6}\) = \(\frac{13}{6}\)
∴ sec θ = \(\frac{13}{12}\)
ఇంకా 2 tan θ = \(\frac{2}{3}\) – \(\frac{3}{2}\) = \(\frac{4-9}{6}\) = \(\frac{-5}{6}\)
∴ tan θ = \(\frac{-5}{12}\)
ఇప్పుడు sin θ = \(\frac{\tan \theta}{\sec \theta}=\frac{\left(\frac{-5}{12}\right)}{\left(\frac{13}{12}\right)}=\frac{-5}{13}\)
∵ tan θ రుణాత్మకం, sec θ ధనాత్మకం.
⇒ θ నాలుగో పాదంలో ఉంటుంది.

ప్రశ్న 6.
cot \(\frac{\pi}{16}\) . cot \(\frac{2\pi}{16}\) . cot \(\frac{3\pi}{16}\) …………. cot \(\frac{7\pi}{16}\) = 1 అని నిరూపించండి.
సాధన:
L.H.S. = \(\frac{\pi}{16}\) . cot \(\frac{2\pi}{16}\) . cot \(\frac{3\pi}{16}\) …………. cot \(\frac{7\pi}{16}\)
= \(\left(\cot \frac{\pi}{16} \cdot \cot \frac{7 \pi}{16}\right)\left(\cot \frac{2 \pi}{16} \cdot \cot \frac{6 \pi}{16}\right)\) \(\left(\cot \frac{3 \pi}{16}, \cot \frac{5 \pi}{16}\right) \cot \left(\frac{4 \pi}{16}\right)\)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 4
= (1) (1) (1) (1) = 1

ప్రశ్న 7.
3 sin θ + 4 cos θ = 5 అయితే 4 sin θ – 3 cos θ విలువ కనుక్కోండి.
సాధన:
3 sin θ + 4 cos θ = 5
4 sin θ – 3 cos θ = x అనుకుందాం
పై సమీకరణాలను వర్గం చేసి, కలిపితే
(3 sin θ + 4 cos θ)2 + (4 sin θ – 3 cos θ)2 = 52 + x2
⇒ 9 sin2 + 16 cos2 θ + 24 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 24 sin θ cos θ = 25 + x2
⇒ 9 + 16 = 25 + x2
⇒ x2 = 0 ⇒ x = 0
∴ 4 sin θ – 3 cos θ = 0

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 8.
cos θ + sin θ = \(\sqrt{2}\) cos θ అయితే cos θ – sin θ = \(\sqrt{2}\) sin θ అని చూపండి. [May ’11]
సాధన:
cos θ + sin θ = \(\sqrt{2}\) cos θ
sin θ = (\(\sqrt{2}\)-1) cos θ
ఇరువైపులా ( \(\sqrt{2}\) + 1) తో గుణిస్తే
⇒ (\(\sqrt{2}\) + 1) sin θ = (\(\sqrt{2}\) + 1) ( \(\sqrt{2}\) – 1) cos θ
⇒ (\(\sqrt{2}\) + 1) sin θ = (2 – 1) cos θ
⇒ \(\sqrt{2}\) sin θ + sin θ = cos θ
∴ cos θ – sin θ = \(\sqrt{2}\) sin θ

ప్రశ్న 9.
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) విలువను కనుక్కోండి,
సాధన:
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ)
= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2 θ)2]
= 2[(sin2 θ+ cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2θ+ cos2 θ)2 – 2 sin2 θ cos2 θ]
= 2[1 – 3 sin2 θ cos2 θ] – 3[1 – 2 sin2 θ cos2 θ]
= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ
= -1

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 10.
(tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ. cosec2 θ అని నిరూపించండి.
సాధన:
(tan θ + cot θ)2
= tan2θ+ cot2 θ + 2 tan θ cot θ
= tan2θ + cot2θ + 2
= (1 + tan2θ) + (1 + cot2 θ)
= sec2 θ + cosec2 θ
ఇంకా sec2 θ. cosec2 = \(\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}\)
= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta \cdot \cos ^2 \theta}\)
= \(\frac{1}{\cos ^2 \theta} \cdot \frac{1}{\sin ^2 \theta}\)
sec2 θ. cosec2

ప్రశ్న 11.
cos θ > 0, tan θ + sin θ = m, tan θ – sin θ = n అయితే m2 – n2 = 4\(\sqrt{mn}\) అని చూపండి.
సాధన:
m = tan θ + sin θ
n = tan θ – sin θ
⇒ m + n = 2 tan θ
m – n = 2 sin θ
(m + n) (m – n) = 4 tan θ sin θ
⇒ m2 – n2 = 4\(\sqrt{\tan ^2 \theta \cdot \sin ^2 \theta}\)
= 4\(\sqrt{\tan ^2 \theta\left(1-\cos ^2 \theta\right)}\)
= 4\(\sqrt{\tan ^2 \theta-\sin ^2 \theta}\)
= 4\(\sqrt{(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)}\)
= 4\(\sqrt{m n}\)
∴ m2 – n2 = 4\(\sqrt{m n}\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 12.
tan 20° = λ అయితే , \(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \cdot \tan 110^{\circ}}\) = \(\frac{1-\lambda^2}{2 \lambda}\) అని నిరూపించండి.
సాధన:
tan 20° = λ
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 5

ప్రశ్న 13.
sin 75°, cos 75°, tan 75°, cot 75° యొక్క విలువలను కనుక్కోండి.
సాధన:
i) sin 75° = sin (45° + 30°)
sin 45°. cos 30° + cos 45°. sin 30°
= \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

ii) cos (75°) = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°
= \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 6
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 7

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 14.
cos A = \(\frac{5}{13}\), sin B = \(\frac{4}{5}\) 0 < A, B < 90° అయితే sin (A + B) విలువ కనుక్కోండి.
సాధన:
0 < A < 90°, cos A = \(\frac{5}{13}\) ⇒ sin A = \(\frac{12}{13}\)
0 < B < 90°, sin B = \(\frac{4}{5}\) ⇒ cos B = \(\frac{3}{5}\)
∴ sin (A + B) = sin A cos B + cos A sin B
= \(\frac{12}{13}\) . \(\frac{3}{5}\) + \(\frac{5}{13}\) . \(\frac{4}{5}\)
= \(\frac{56}{65}\)

ప్రశ్న 15.
sin2\(\left(52 \frac{1}{2}\right)^{\circ}\) – sin2\(\left(22 \frac{1}{2}\right)^{\circ}\) = \(\frac{\sqrt{3}+1}{4 \sqrt{2}}\) అని నిరూపించండి.
సాధన:
సూచన : sin2 A – sin2 B = sin (A + B). sin (A – B)
∴ sin2 52\(\frac{1}{2}\) – sin2 22\(\frac{1}{2}\)
= sin (52\(\frac{1}{2}\)° +22\(\frac{1}{2}\)°) . sin (52\(\frac{1}{2}\)° – 22\(\frac{1}{2}\)°)
= sin (75°). sin 30°
= \(\frac{1}{2}\) sin (45° + 30°)
= \(\frac{1}{2}\) [sin 45° cos 30° + cos 45° sin 30°]
= \(\frac{1}{2}\left[\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right]\)
= \(\frac{\sqrt{3}+1}{4 \sqrt{2}}\)

ప్రశ్న 16.
tan 70° – tan 20° = 2 tan 50° అని నిరూపించండి.
సాధన:
tan 50° = tan (70° – 20°)
= \(\frac{\tan 70^{\circ}-\tan 20^{\circ}}{1+\tan 20^{\circ} \tan 70^{\circ}}\)
⇒ tan 70° – tan 20°
= tan 50° [1 + tan 20° tan (90° – 20°)]
⇒ tan 70° – tan 20° = tan 50° [1 + tan 20° cot 20°]
⇒ tan 70° – tan 20° = 2 tan 50°

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 17.
A + B = π/4 అయితే ,
i) (1 + tan A) (1 + tan B) = 2, [Mar. 11, ’07]
ii) (cot A – 1) (cot B – 1) = 2 అని నిరూపించండి.
సాధన:
i) A + B = π/4
⇒ tan (A + B) = tan (π/4)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
ఇరువైపులా ‘1’ కలుపగా
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
⇒ (1 + tan A) (1 + tan B) = 2

ii) A + B = π/4
⇒ cot (A + B) = cot π/4
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = 1
⇒ cot A cot B – 1 = cot B + cot A
⇒ cot A cot B – cot A – cot B = 1
ఇరువైపులా ‘1’ కలుపగా
cot A cot B – cot A – cot B + 1 = 1 + 1
∴ (cot A – 1) (cot B – 1) = 2.

ప్రశ్న 18.
sin α = \(\frac{1}{\sqrt{10}}\), sin β = \(\frac{1}{\sqrt{5}}\), α, β లము కోణాలయితే α + β = π/4 అని చూపండి.
సాధన:
α ఒక లఘు కోణం, sin α = \(\frac{1}{\sqrt{10}}\) ⇒ tan α = \(\frac{1}{2}\)
β ఒక లఘు కోణం, sin β = \(\frac{1}{\sqrt{5}}\) ⇒ tan β = \(\frac{1}{2}\)
కనుక tan (α + β) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)
= \(\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3} \cdot \frac{1}{2}}\) = 1 ⇒ α + β = π/4

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 19.
sin A = \(\frac{12}{13}\), cos B = \(\frac{3}{5}\), A, B రెండూ మొదటి పాదంలో లేకపోతే, A + B ఏ పాదంలో ఉందో కనుక్కోండి.
సాధన:
sin A = \(\frac{12}{13}\), A మొదటి పాదంలో లేదు.
⇒ A రెండవ పాదంలో ఉంటుంది ∵ sin A +ve
cos B = \(\frac{3}{5}\), θ మొదటి పాదంలో లేదు.
⇒ B నాల్గవ పాదంలో ఉంటుంది ∵ cos B +ve
∵ sin A = \(\frac{12}{13}\) ⇒ cos A = \(\frac{-5}{13}\)
cos B = \(\frac{3}{5}\) ⇒ sin B = \(\frac{-4}{5}\)
sin (A+B) sin A cos B + cos A sin B
= \(\left(\frac{12}{13}\right)\left(\frac{3}{5}\right)+\left(\frac{-5}{13}\right)\left(\frac{-4}{5}\right)\)
= \(\frac{36+20}{65}\) = \(\frac{56}{65}\)
cos (A+B)= cos A cos B – sin A sin B.
= \(\left(\frac{-5}{13}\right)\left(\frac{3}{5}\right)-\left(\frac{12}{13}\right)\left(\frac{-4}{5}\right)\)
= \(\frac{-15+48}{65}\) = \(\frac{33}{65}\)
∵ sin (A + B), cos (A + B) లు రెండూ ధనాత్మకాలు
⇒ (A + B) మొదటి పాదంలో ఉంటుంది.

ప్రశ్న 20.
(i) tan A పదాలలో tan \(\left(\frac{\pi}{4}+\mathrm{A}\right)\) ని,
(ii) cot A పదాలలో cot \(\left(\frac{\pi}{4}+\mathrm{A}\right)\) ని కనుక్కోండి.
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 8

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 21.
\(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = cot 36° అని నిరూపించండి. [(A.P) Mar. 15, ’11
సాధన:
L.H.S. = \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\)
లవ, హారాలను cos 9° తో భాగిస్తే,
= \(\frac{1+\tan 9^{\circ}}{1-\tan 9^{\circ}}\) [∵ tan \(\left(\frac{\pi}{4}+\theta\right)=\frac{1+\tan \theta}{1-\tan \theta}\)]
= tan (45° + 9°)
= tan (54°)
= tan (90° – 36°) = cot 36°
∴ \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = cot 36°.

ప్రశ్న 22.
cos 42° + cos 78° + cos 162° 0 అని చూపండి. [May ’11]
సాధన:
L.H.S.= cos 42° + cos 78° + cos 162°
= 2 cos \(\left(\frac{42^{\circ}+78^{\circ}}{2}\right)\) . cos \(\left(\frac{42^{\circ}-78^{\circ}}{2}\right)\) + cos (180° – 18°)
= 2 cos 60°. cos (-18°) + cos (180° – 18°)
= 2\(\left(\frac{1}{2}\right)\) cos 18° – cos 18° = 0
∴ cos 42° + cos 78° + cos 162°
= 0

ప్రశ్న 23.
\(\sqrt{3}\) sin θ + cos θ ని ఒక కోణానికి sine. గా వ్యక్తపరచండి.
సాధన:
\(\sqrt{3}\) sin θ + cos θ = 2(\(\frac{\sqrt{3}}{2}\) sinθ + \(\frac{1}{2}\)cos θ)
= 2(cos \(\frac{\pi}{6}\) sin 0 + sin \(\frac{\pi}{6}\) cos θ)
= 2. sin (θ + \(\frac{\pi}{6}\)).

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 24.
sin2 θ + sin2 [θ – \(\frac{\pi}{3}\)] + sin2 [θ – \(\frac{\pi}{3}\)] = \(\frac{3}{2}\) అని చూపండి.
సాధన:
LHS = sin2 θ + sin2 [θ – \(\frac{\pi}{3}\)] + sin2 [θ – \(\frac{\pi}{3}\)]
= sin2 θ + (sin θ cos \(\frac{\pi}{3}\) + cos θ sin \(\frac{\pi}{3}\))2 + (sin θ cos cos \(\frac{\pi}{3}\) – cos \(\frac{\pi}{3}\) sin θ)2
= sin2 θ + 2 (sin2 θ cos2 \(\frac{\pi}{3}\) + cos2 θ sin2\(\frac{\pi}{3}\))
= sin2 θ + 2 (sin2 θ . \(\frac{1}{4}\) + cos2θ . \(\frac{3}{4}\))
= sin2 θ + \(\frac{1}{2}\) sin2 θ + \(\frac{3}{2}\) cos2 θ
= \(\frac{3}{2}\) sin2 θ + \(\frac{3}{2}\) cos2 θ = \(\frac{3}{2}\) (sin2 θ + cos2 θ)
= \(\frac{3}{2}\) = R.H.S.

ప్రశ్న 25.
A, B, C లు ఒక త్రిభుజంలోని కోణాలు అవుతూ, వీటిలో ఏ ఒక్కటీ \(\frac{\pi}{2}\) కానప్పుడు
i) tan A+ tan B + tan C = tan A tan B tan C
ii) cot A cot B+ cot B cot C + cot C cot A = 1 అని నిరూపించండి.
సాధన:
i) A, B, C లు త్రిభుజ కోణాలు.
⇒ A + B + C = 180°
⇒ A + B = 180° – C
⇒ tan (A + B) = tan (180° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
⇒ tan A + tan B = -tan C + tan A tan B tan C
⇒ tan A + tan B + tan C = tan A tan B tan C

ii) A + B + C = 180°
⇒ A+B 180° – C
⇒ cot (A + B) = cot (180° – C)
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = -cot C
⇒ cot A cot B – 1 = -cot B cot C – cot C. cot A
⇒ cot A cot B + cot B cot C + cot C cot A = 1

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 26.
ABC త్రిభుజంలో cot A + cot B + cot C = \(\sqrt{3}\) అయితే, ABC ఒక సమబాహు త్రిభుజమని చూపండి.
సాధన:
A + B + C = 180°
పై లెక్క నుండి
cot A cot B + cot B cot C+ cot C cot A = 1
i.e., Σ (cot A cot B) = 1 ………………. (1)
ఇప్పుడు (cot A – cot B)2
= Σ (cot2 A + cot2 B – 2 cot A cot B)
= 2 cot2 A + 2 cot2 B + 2 cot2 C – 2 cot A cot B – 2 cot B cot C – 2 cot C cot A
= 2 [cot A+ cot B + cot C]2 – 6 (cot A cot B + cot B cot C + cot C cot A)
= 2|(\(\sqrt{3}\))2 | – 6(1)
= 6 – 6 = 0
∴ Σ (cot A – cot B)2 = 0
⇒ cot A = cot B = cot C
⇒ cot A = cot B = cot C = \(\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)
(∵ cot A + cot B + cot C = \(\sqrt{3}\))
⇒ A = B = C = 60°
∴ ∆ABC సమబాహు త్రిభుజం

ప్రశ్న 27.
x = tan A, y tan B, z = tan C, A, B, C, A – B, B – C, C – A లలో ఏ ఒక్కటి కూడ \(\frac{\pi}{2}\) బేసి గుణిజం కానట్లయితే \(\Sigma\left(\frac{x-y}{1+x y}\right)=\pi\left(\frac{x-y}{1+x y}\right)\) అని చూపండి.
సాధన:
∵ x = tan A, y = tan B, z = tan C
అప్పుడు. \(\frac{x-y}{1+x y}=\frac{\tan A-\tan B}{1+\tan A \tan B}\) = tan (A – B)
ఇదే విధంగా \(\frac{y-z}{1+y z}\) = tan (B – C) మరియు
\(\frac{z-x}{1+z x}\) = tan (C – A)
P = A – B, Q = B – C, R = C – A అని రాస్తే
P + Q + R = 0 కనుక
⇒ P + Q = R
⇒ tan (P + Q) = tan (-R)
⇒ \(\frac{\tan P+\tan Q}{1-\tan P \tan Q}\) = -tan R
⇒ tan P + tan Q = -tan R + tan P tan Q tan R
⇒ tan P + tan Q + tan R = tan P tan Q tan R
⇒ Σ (tan P) = π (tan P)
⇒Σ tan (A – B) = π tan (A – B)
∴ \(\Sigma\left(\frac{x-y}{1+x y}\right)=\pi\left(\frac{x-y}{1+x y}\right)\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 28.
i) sin 22\(\frac{1}{2}\)°
ii) cos 22\(\frac{1}{2}\)°
iii) tan 22\(\frac{1}{2}\)°
iv) cot 22\(\frac{1}{2}\)° విలువలను కనుక్కోండి.
సాధన:
A = 2222\(\frac{1}{2}\)° అయిన 2A = 45° అవుతుంది.
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 9
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 10

ప్రశ్న 29.
కింది విలువలను కనుక్కోండి.
i) sin 67\(\frac{1}{2}\)°
ii) cos 67\(\frac{1}{2}\)°
iii) tan 67\(\frac{1}{2}\)°
iv) cot 67\(\frac{1}{2}\)°
సాధన:
A = 67\(\frac{1}{2}\)° అయిన 2A = 135° అవుతుంది.
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 11
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 12

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 30.
\(\frac{1-\cos 2 \theta}{\sin 2 \theta}\) సూక్ష్మీకరించండి.
సాధన:
\(\frac{1-\cos 2 \theta}{\sin 2 \theta}=\frac{2 \sin ^2 \theta}{2 \sin \theta \cos \theta}=\frac{\sin \theta}{\cos \theta}\) = tan θ.

ప్రశ్న 31.
cos A = \(\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\) అయితే cos 2A విలువను కనుక్కోండి.
సాధన:
cos 2A = 2cos2 A – 1 = 2 \(\left(\frac{\sqrt{2}+1}{2 \sqrt{2}}\right)\) – 1
= \(\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\) – 1
= \(\frac{1}{\sqrt{2}}\)

ప్రశ్న 32.
\(\frac{\pi}{2}\) < 0 < π, cos θ = \(\frac{-5}{13}\) అయితే sin 2θ విలువను కనుక్కోండి.
సాధన:
\(\frac{\pi}{2}\) < θ < π ⇒ sin θ > 0, cos θ = –\(\frac{5}{13}\)
⇒ sin θ = \(\frac{12}{13}\)
∴ sin 2θ = 2 sin θ.cos θ
= 2 . \(\frac{12}{13}\left(\frac{-5}{13}\right)\) = \(\frac{-120}{169}\)

ప్రశ్న 33.
ప్రథమ పాదంలోని x యొక్క ఏ విలువలకు \(\frac{2 \tan x}{1-\tan ^2 x}\) ధనాత్మకం.
సాధన:
\(\frac{2 \tan x}{1-\tan ^2 x}\) > 0 ⇒ tan 2x > 0
⇒ 0 < 2x < \(\frac{\pi}{2}\) (x మొదటి పాదంలోని కోణం కనక)
⇒0 < x < \(\frac{\pi}{4}\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 34.
π < θ < \(\frac{3\pi}{2}\), cos θ = \(\frac{-3}{5}\) అయితే tan θ/2 విలువను కనుక్కోండి.
సాధన:
tan \(\frac{\theta}{2}=\pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\pm \sqrt{\frac{1+\frac{3}{5}}{1-\frac{3}{5}}}\) = ±2
దత్తాంశం నుంచి π < θ < \(\frac{3\pi}{2}\) ⇒ \(\frac{\pi}{2}\) < \(\frac{\theta}{2}\) < \(\frac{3\pi}{4}\)
⇒ tan θ/2 < 0
∴ tan θ/2 = -2

ప్రశ్న 35.
A అనేది \(\frac{\pi}{2}\) పూర్ణాంక గుణిజం కాకపోతే
i) tan A + cot A = 2 cosec 2A
ii) cot A tan A = 2 cot 2A
సాధన:
i) tan A + cot A = \(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 13
∴ tan A + cot A = 2 cosec 2A

ప్రశ్న 36.
θ అనేది \(\frac{\pi}{2}\) పూర్ణాంక గుణిజం కాకపోతే, tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ అని నిరూపించండి.
సాధన:
cot A – tan A = 2 cot 2A
tan A = cot A – 2 cot 2A అని వస్తుంది ……………… (1)
కనుక , tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ
= (cot θ – 2 cot 2θ) + 2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ
(సమీకరణం (1) నుంచి)
cot θ

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 37.
A ∈ R అయితే,
i) sin A sin (π/3 + A) sin (π/3 – A) = \(\frac{1}{4}\) sin 3A
ii) cos A. cos (π/3 + A) cos (π/3 – A) = \(\frac{1}{4}\) cos 3A అని నిరూపించి తద్వారా
iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
iv) cos \(\frac{\pi}{9}\) cos \(\frac{2\pi}{9}\) cos \(\frac{3\pi}{9}\) cos \(\frac{4\pi}{9}\) = \(\frac{1}{16}\) అని రాబట్టండి.
సాధన:
i) sin A. sin (π/3 + A). sin (π/3 – A)
sin A [sin2 π/3 sin2 A]
= [sin (A + B). sin (A – B)
= sin2 A – sin2 B = cos2 B – cos2 A]
= sin A \(\left[\left(\frac{3}{2}\right)-\sin ^2 \mathrm{~A}\right]\)
= sin A \(\left(\frac{3-4 \sin ^2 A}{4}\right)\)
= \(\frac{1}{4}\) (3 sin A – 4 sin3 A)
= \(\frac{1}{4}\) |sin 3A|

ii) cos A cos (π/3 + A) cos (π/3 – A)
= cos A (cos2 A – sin2 \(\frac{\pi}{3}\)_
= cos A (cos2 A – \(\frac{3}{4}\))
= \(\frac{\cos A\left(4 \cos ^2 A-3\right)}{4}\)
= \(\frac{1}{4}\) (4 cos3 A – 3 cos A)
= \(\frac{1}{4}\) . cos 3A
= \(\frac{1}{4}\) [4 cos3 A – 3 cos A]
= \(\frac{1}{4}\) cos 3A

iii) (1) A 20° ప్రతిక్షేపిస్తే,
⇒ sin 20°. sin (60° + 20°) sin (60° – 20°) = \(\frac{1}{4}\) sin 3(20°)
లేదా sin 20° sin 80° sin 40° = \(\frac{1}{4}\) sin 60°
పై సమీకరణాన్ని ఇరువైపులా sin 60° తో గుణిస్తే
sin 20° sin 40° sin 60° sin 80°
= \(\frac{1}{4}\) sin2 60° = \(\frac{1}{4}\) \(\frac{3}{4}\) = \(\frac{3}{16}\)
ఇదేవిధంగా, (ii) నుంచి

iv) cos 20° cos 40° cos 80° = \(\frac{1}{4}\) . cos 60° అని వస్తుంది.
ఇరువైపులా cos 60° తో గుణిస్తే
cos 20° cos 40° cos 60° cos 80°
= \(\frac{1}{4}\) cos2 60° = \(\frac{1}{16}\)
అంటే cos \(\frac{\pi}{9}\) cos \(\frac{2\pi}{9}\) cos \(\frac{3\pi}{9}\) cos \(\frac{4\pi}{9}\) = \(\frac{1}{16}\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 38.
3A అనేది \(\frac{\pi}{2}\) కి బేసి గుణిజం కాకపోతే, tan A. tan(60° + A). tan (60° – A) = tan 3A అని నిరూపించి తర్వాత tan 6° tan 42° tan 66° tan 78° విలువను కనుక్కోండి.
సాధన:
sin A sin (π/3 + A) sin (π/3 – A) = 1/4 sin 3A
cos A cos (π/3 + A) cos (π/3 – A) = 1/4 cos 3A
sin A. sin (60° + A). sin (60° – A) = \(\frac{1}{4}\) sin 3A …………… (1)
cos A.cos (60° + A). cos (60° – A) = \(\frac{1}{4}\) cos 3A …………… (2)
(1) ని (2) తో భాగిస్తే
tan A. tan (60° + A). tan (60° – A) = tan 3A …………….. (3)
(3) లో A = 6° ప్రతిక్షేపిస్తే
tan 6°. tan 66°. tan 54° = tan 18° …………… (4)
మళ్ళీ (3) లో A = 18° ప్రతిక్షేపిస్తే
tan 18°. tan 78°. tan 42° = tan 54° …………… (5)
(4), (5) సమీకరణాలను గుణిస్తే
(tan 6°. tan 66°. tan 54°).
(tan 18°. tan 78°. tan 42°) = tan 18°. tan 54°
కనుక tan 6°. tan 42°. tan 66°. tan 78° = 1.

ప్రశ్న 39.
α, β ∈ R అయితే , (cos α + cos β)2 + (sin α + sin β)2 = 4 cos2 \(\frac{(\alpha+\beta)}{2}\) అని నిరూపించండి.
సాధన:
L.H.S. = (cos α + cos β)2 + (sin α + sin β)2
= (cos2 α + cos2 β + 2 cos α cos β) + (sin2 α + sin2 β + 2 sin α sin β)
= 1 + 1 + 2 (cos α cos β + sin α sin β)
= 2 + 2 cos (α – β)
= 2 [1 + cos (α – β)]
= 2[2 cos2 \(\frac{(\alpha+\beta)}{2}\)]
= 4. cos2 \(\frac{(\alpha+\beta)}{2}\) [∵ 1+ cos θ = 2 cos2 = \(\frac{\theta}{2}\)]

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 40.
a, b, c లు శూన్యేతర వాస్తవ సంఖ్యలు, a cos θ + b sin θ = c సమీకరణానికి సాధనలు α, β లు అయితే
(i) sin α + sin β = \(\frac{2 b c}{a^2+b^2}\)
(ii) sin α . sin β = \(\frac{c^2-a^2}{a^2+b^2}\) అని చూపిండి.
సాధన:
∵ a cos θ + b sin θ = c
⇒ a cos θ = c – b sin θ
⇒ (a cos θ)2 = (c – b sin θ)2
⇒ a2 cos2 θ = c2 + b2 sin2 θ – 2bc sin θ
⇒ a2 (1 – sin2 θ) = c2 + b2 sin2 θ – 2bc sin θ
⇒ (a2 + b2) sin2 θ – 2bc sin θ + (c2 – a2) = 0
ఇది sin θ లో వర్గ సమీకరణం. sin α, sin β లు దీనికి మూలాలు
∴ మూలాలు మొత్తం = sin α + sin β = \(\frac{2 b c}{a^2+b^2}\)
మూలాల లబ్దం = sin α sin β = \(\frac{c^2-a^2}{a^2+b^2}\)

ప్రశ్న 41.
θ అనేది \(\frac{\pi}{2}\) బేసి గుణిజం కాదు, cos θ ≠ \(\frac{-1}{2}\) అయినప్పుడు, \(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}\) = tan θ అని నిరూపించండి.
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 14

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 42.
sin4 \(\frac{\pi}{8}\) + sin4 \(\frac{3\pi}{8}\) + sin4 \(\frac{5\pi}{8}\) + sin4 \(\frac{7\pi}{8}\) = \(\frac{3}{2}\) అని నిరూపించండి.
సాధన:
L.H.S. = sin4 \(\frac{\pi}{8}\) + sin4 \(\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\) + sin4 \(\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\) + sin4 \(\left(\pi-\frac{\pi}{8}\right)\)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 15

ప్రశ్న 43.
2A, 3A లు \(\frac{\pi}{2}\) బేసి గుణిజాలు కానప్పుడు, tan 3A. tan 2A. tan A = tan 3A – tan 2A – tan A అని రుజువు చేయండి.
సాధన:
tan 3A = tan (2A + A)
= \(\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
⇒ tan 3A(1 – tan 2A tan A) = tan 2A + tan A
⇒ tan 3A – tan A tan 2A tan 3A = tan 2A + tan A
∴ tan 3A – tan 2A – tan A = tan A tan 2A tan 3A

ప్రశ్న 44.
sin 78° + cos 132° = \(\frac{\sqrt{5}-1}{4}\) అని చూపండి.
సాధన:
L.H.S. = sin 78° + cos 132°
= sin 78° + cos (90° + 42°)
= sin 78° – sin 42°
= 2 cos\(\left(\frac{78^{\circ}+42^{\circ}}{2}\right)\) . sin \(\left(\frac{78^{\circ}-42^{\circ}}{2}\right)\)
= 2 cos (60°), sin (18°)
= 2\(\left(\frac{1}{2}\right)\left(\frac{\sqrt{5}-1}{4}\right)\) = \(\frac{\sqrt{5}-1}{4}\) = R.H.S.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 45.
sin 21° cos 9° cos 84° cos 6° = \(\frac{1}{4}\) అని చూపండి.
సాధన:
L.H.S. = sin 21° cos 9° cos 84° cos 6°
= \(\frac{1}{4}\) [2 sin 21° cos 9° 2 cos 84° cos 6°]
= \(\frac{1}{2}\) [sin (21° + 9°) + sin (21° – 9°) – 2 cos (90° – 6°) cos 6°]
= \(\frac{1}{2}\) [sin 30°+ sin 12° – 2 sin 6° cos 6°]
= \(\frac{1}{2}\) [\(\frac{1}{4}\) + sin 12° – sin (2 × 16°)
= \(\frac{1}{4}\) = R.H.S.

ప్రశ్న 46.
sin 34° + cos 64° – cos 4° విలువను కనుక్కోండి .
సాధన:
sin 34° + (cos 64° – cos 4°)
= sin 34° – 2 sin \(\frac{64^{\circ}+4^{\circ}}{2}\) sin \(\frac{64^{\circ}-4^{\circ}}{2}\) [cos C – cosD = – 2 sin \(\left(\frac{C+D}{2}\right)\) sin \(\left(\frac{C-D}{2}\right)\)]
= sin 34° – 2. sin 34°.sin 30°
= 0 (sin 30° = \(\frac{1}{2}\) కనుక)

ప్రశ్న 47.
cos2 76° + cos2 16° – cos 76°. cos 16° = \(\frac{3}{4}\) అని చూపండి.
సాధన:
L.H.S.= cos2 76° + cos2 16° – cos 76° cos 16°
= cos2 76° + (1 – sin2 16°) – \(\frac{1}{2}\) (2 cos 76° cos 16°)
= 1 + (cos2 76° – sin2 16°) – \(\frac{1}{2}\) [cos (76° + 16°) + cos (76° – 16°)]
= 1 + cos (76° +16°). cos (76° -16°) – \(\frac{1}{2}\) [cos 92° + cos 60°]
= 1 + cos (92°). cos 60° – \(\frac{1}{2}\) cos 92° – \(\frac{1}{2}\) cos 60°
= 1 + \(\frac{1}{2}\) . cos 92° – \(\frac{1}{2}\) cos 92° – \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
= R.H.S.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 48.
a ≠ 0b, sin x + sin y = a, cos x + cos y = b అయితే
i) tan \(\frac{x+y}{2}\)
ii) sin \(\frac{x-y}{2}\) విలువలు a, b లలో కనుక్కోండి.
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 16

మొదటి పద్ధతి :
ii) a2 + b2 = (sin x + sin y)2 + (cos x + cos y)2
= sin2 x + sin2 y + 2 sin x sin y + cos2 x + cos2 y + 2 cos x cos y
= 2 + 2(cos x cos y + sin x sin y)
= 2 + 2 cos(x – y)
= 2[1 + cos(xy)]
a2 + b2 – 2 = 2 cos(x – y)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 17
రెండవ పద్ధతి:
a2 + b2 = (sin x + sin y)2 + (cos x + cos y)2
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 18

ప్రశ్న 49.
cos 12° + cos 84° + cos 132° + cos 156° = –\(\frac{1}{2}\) అని చూపండి.
సాధన:
L.H.S.= cos 12° + cos 84° + cos 132° + cos 156°
= (cos 132°+ cos 12°) + (cos 156° + cos 84°)
= 2 cos \(\left(\frac{132^{\circ}+12^{\circ}}{2}\right)\) . cos \(\left(\frac{132^{\circ}-12^{\circ}}{2}\right)\) + 2 cos \(\left(\frac{156^{\circ}+84^{\circ}}{2}\right)\) . cos \(\left(\frac{156^{\circ}-84^{\circ}}{2}\right)\)
= 2 cos 72°. cos 60° + 2 cos (120°). cos (36°)
= 2 cos (90° – 18°)\(\left(\frac{1}{2}\right)\) + 2\(\left(\frac{-1}{2}\right)\) cos 36°
= sin 18° – cos 36°
= \(\left(\frac{\sqrt{5}-1}{4}\right)-\left(\frac{\sqrt{5}+1}{4}\right)=\frac{-1}{2}\) = R.H.S

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 50.
ప్రతీ θ ∈ R కి, 4 sin \(\frac{5 \theta}{2}\) cos \(\frac{2 \theta}{2}\) cos 3θ = sin θ – sin 2θ + sin 4θ + sin 7θ అని చూపండి.
సాధన:
L.H.S. = 4 sin \(\frac{5 \theta}{2}\) cos \(\frac{2 \theta}{2}\) cos 3θ
= 2 [2 sin \(\frac{5 \theta}{2}\) cos \(\frac{3 \theta}{2}\) . cos 3θ
= 2 [sin\(\left(\frac{5 \theta}{2}+\frac{3 \theta}{2}\right)\) + sin \(\left(\frac{5 \theta}{2}-\frac{3 \theta}{2}\right)\)] cos 3θ
= 2[(sin 4θ + sin θ) cos 3θ]
= 2 sin 4θ cos 3θ + 2 sin θ cos 3θ
= sin (4θ + 3θ) + sin (4θ – 3θ) + sin (θ + 3θ) + sin (θ – 3θ)
= sin 7θ + sin θ + sin 4θ – sin 2θ = R.H.S.

ప్రశ్న 51.
A, B, A + B లలో ఏదీ π కి పూర్ణాంక గుణిజం కాకపోతే, \(\frac{1-\cos A+\cos B-\cos (A+B)}{1+\cos A-\cos B-\cos (A+B)}\) = tan \(\frac{A}{2}\) cot \(\frac{A}{2}\) అని చూపండి.
సాధన:
లవం = 1 – cos A + cos B cos (A + B)
= [1 – cos (A + B)] – (cos A – cos B)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 19
(1), (2) ల నుండి
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 20

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 52.
α ∈ R అయితే , cos2 (α – π/4) + cos2 (α + π/12) – cos2 (α – π/12) = \(\frac{1}{2}\) అని చూపండి.
సాధన:
L.H.S. = cos2 (α – π/4) + cos2 (α + π/12) – cos2(α – π/2)
= \(\frac{1+\cos 2(\alpha-\pi / 2)}{2}\) + {sin (α + π/12 + α – π/12) sin (α – π/12 – α – π/12)}
= \(\frac{1}{2}\) (1 + sin 2α) + sin 2α. sin (π/6)
= \(\frac{1}{2}\)+ \(\frac{1}{2}\) sin 2α – \(\frac{1}{2}\) sin 2α = \(\frac{1}{2}\)
సూచన : cos2 A – cos2 B = sin (A + B) sin (B – A)

ప్రశ్న 53.
(α – β) అనేది \(\frac{\pi}{2}\) బేసి గుణిజం కాదు, m ≠ – 1, \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\dot{\beta})}=\frac{1-m}{1+m}\) అయితే tan \(\left(\frac{\pi}{4}-\alpha\right)\) = m. tan \(\left(\frac{\pi}{4}+\beta\right)\) అని చూపండి.
సాధన:
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 21
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 22
∴ tan \(\left(\frac{\pi}{4}-\alpha\right)\) = m. tan \(\left(\frac{\pi}{4}+\beta\right)\)

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 54.
A, B, C లు త్రిభుజ కోణాలయితే, కింది వాటిని నిరూపించండి.
i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
ii) sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C
సాధన:
i) ∵ A, B, C లు త్రిభుజ కోణాలు
⇒ A + B + C = 180° ……………… (1)
L.H.S. = sin 2A + sin 2B + sin 2C
= 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) . cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + sin 2C
= 2 sin (A + B) cos (AB) + sin 2C
= 2 sin (180° C) cos (AB) + sin 2C
(1) నుండి
2 sin C. cos (A – B) + 2 sin C. cos C
= 2 sin C [cos (A – B) + cos C]
= 2 sin C [cos (A – B) + cos (180° – \(\overline{A+B}\))]
= 2 sin C [cos (AB) cos (A + B)] = 2 sin C (2 sin A sin B)
= 4 sin A sin B sin C = R.H.S.

ii) L.H.S. = sin 2A + sin 2B – sin 2C
= 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) . cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) – sin 2C
= 2 sin (A + B) cos (A – B) sin 2C
= 2 sin (180° – C) cos (A – B) – sin 2C
= 2 sin C. cos (A – B) – 2 sin C cos C
= 2 sin C [cos (A – B) cos C]
= 2 sin C [cos (A – B) – cos (180° – \(\overline{A+B}\))]
(1) నుండి
= 2 sin C [cos (A – B) + cos (A + B)]
= 2 sin C [2 cos A cos B]
= 4 cos A cos B sin C = R.H.S.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 55.
A, B, C లు త్రిభుజ కోణాలైతే, కింది వాటిని నిరూపించండి.
i) cos 2A + cos 2B + cos 2C = -4 cos A cos B cos C – 1
ii) cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin B cos C
సాధన:
i) ∵ A, B, C లు త్రిభుజ కోణాలు.
⇒ A + B + C = 180° ……………. (1)
L.H.S.= (cos 2A + cos 2B) + cos 2C
= 2 cos (A + B) cos (A – B) + 2 cos2 C – 1
= -2 cos C cos (A – B) – 2 cos C cos (A + B) – 1
[∵ cos (A + B) = -cos C]
= -1 – 2 cos C [cos (AB) + cos (A – B)]
= -1 – 2 cos C [2 cos A cos B]
= -1 – 4 cos A cos B cos C = R.H.S.

ii) L.H.S.= cos 2A + cos 2B – cos 2C
= 2 cos \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) – cos 2C
= 2 cos (A + B). cos (A – B) – (2 cos2 C – 1)
= 2 cos (180° – C) cos (A – B) – 2 cos2 C + 1
= 1 – 2 cos C cos (A – B) – 2 cos2 C
= 1 – 2 cos C [cos (AB) + cos C]
= 1 – 2 cos C [cos (A – B) + cos (180° \(\overline{A+B}\))]
= 1 – 2 cos C [cos (AB) cos (A + B)]
= 1 – 2 cos C [2 sin A sin B]
= 1 – 4 sin A sin B cos C = R.H.S.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 56.
A, B, C లు త్రిభుజ కోణాలయితే కింది వాటిని నిరూపించండి.
i) sin A + sin B + sin C = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)
ii) cos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)
సాధన:
i) ∵ A, B, C లు త్రిభుజ కోణాలు.
⇒ A + B + C = 180°
L.H.S. = (sin A+ sin B) + sin C
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 23

ii) cos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)
L.H.S. = cos A + cos B + cos C
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 24
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 25

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 57.
A + B + C = 9 π/2 0° అయితే ,
i) sin2 A + sin2 B + sin2 C = 1 – 2 sin A sin B sin C
ii) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C అని చూపండి.
సాధన:
i) A + B + C = π/2 ………………. (1)
L.H.S. = sin2 A + sin2 B + sin2 C
= \(\frac{1}{2}\) [1 – cos 2A + 1 – cos 2B + 1 – cos 2C]
= \(\frac{1}{2}\) [3 – (cos 2A + cos 2B + cos 2C)]
= \(\frac{1}{2}\) [3 – (1 + 4 sin A sin B sin C)]
(56(ii)వ లెక్క చూడండి.)
(∵ 2A + 2B + 2C = 180°)
= \(\frac{1}{2}\) [2 – 4 sin A sin B sin C]
= 1 – 2 sin A sin B sin C

ii) A + B + C = π/2
⇒ 2A + 2B + 2C = 180°
sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
(56(i)వ లెక్క చూడండి.)

ప్రశ్న 58.
A + B + C = 3π/2, cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C అని చూపండి. [Mar. 13]
సాధన:
A + B + C = 3π/2 …………….. (1)
L.H.S.= cos 2A + cos 2B + cos 2C
= 2 cos (A + B). cos (A – B) + 1 – 2 sin2C
= 2 cos (270° – C). cos (A – B) + 1 – 2 sin2C
= 1 – 2 sin C cos (A – B) – 2 sin2 C
= 1 – 2 sin C [cos (A – B) + sin C]
= 1 – 2 sin C [cos (A – B) + sin (270° – \(\overline{A+B}\))]
= 1 – 2 sin C [cos (A – B) – cos (A + B)]
= 1 – 2 sin C [2 sin A sin B]
= 1 – 4 sin A sin B sin C

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 59.
A, B, C లు త్రిభుజ కోణాలయితే, sin2 \(\frac{A}{2}\) + sin2 \(\frac{B}{2}\) – sin2 \(\frac{C}{2}\) = 1 – 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\) అని చూపండి. [May ’11]
సాధన:
A + B + C = 180° …………… (1)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 26

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 60.
A, B, C లు త్రిభుజ కోణాలయితే sin\(\frac{A}{2}\) + sin\(\frac{B}{2}\) + sin\(\frac{C}{2}\) = 1 + 4 sin \(\frac{\pi-A}{4}\) . sin \(\frac{\pi-B}{4}\) sin \(\frac{\pi-C}{4}\) అని రుజువు చేయండి. [Mar. ’14]
సాధన:
A + B + C = 180° …………….. (1)
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 27
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 28
sin\(\frac{A}{2}\) + sin\(\frac{B}{2}\) + sin\(\frac{C}{2}\) = L.H.S

ప్రశ్న 61.
A + B + C = 0 అయితే, cos2 A + cosఅయితే B + cosఅయితే C = 1 + 2 cos A cos B cos C అని రుజువు చేయండి.
సాధన:
A + B + C = 0 ………….. (1)
L.H.S. = cos2 A + cos2 B + cos2 C
= cos2 A + (1 – sin2 B) + cos2 C
= 1 + (cos2 A – sin2 B) + cos2 C
= 1 + cos (A + B) cos (A – B) + cos2 C
= 1 + cos (-C) cos (A – B) + cos2 C
(1) నుండి
= 1 + cos C cos (A – B) + cos2 C
= 1 + cos C [cos (A – B) + cos C]
= 1 + cos C [cos (A – B) + cos (- B – A)]
(1) నుండి
= 1 + cos C [cos (A – B) + cos (A + B)]
= 1 + cos C (2 cos A cos B)
= 1 + 2 cos A cos B cos C
= R.H.S.

AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు

ప్రశ్న 62.
A + B + C = 2S అయితే cos (S – A) + cos (S – A) + cos (S – B) + cos (S – C) + cos S = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\) అని రుజువు చేయండి.
సాధన:
A + B + C = 2S ……………. (1)
L.H.S. = [cos (S – A) + cos (S – B) + cos (S – C) + cos S]
AP Inter 1st Year Maths 1A Important Questions Chapter 6 త్రికోణమితీయ నిష్పత్తులు, పరివర్తనలు 29

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 2nd Lesson ప్రమాణాలు, కొలత Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 2nd Lesson ప్రమాణాలు, కొలత

అతిస్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
యధార్థత, ఖచ్చితత్వాల మధ్య తేడాను రాయండి. [May, Mar. ’13]
జవాబు:

యధార్థతఖచ్చితత్వం
1) మనం కొలిచే భౌతిక రాశి నిజమైన విలువకు ఎంత దగ్గరగా. ఉన్నది తెలియజేసే ఒక కొలమానమే మనం తీసుకున్న కొలత యొక్క యధార్ధత.1) ఖచ్చితత్వం అనేది, ఆ పరికరంతో మనం ఎంత కనిష్ఠ అవధి వరకు (లేదా) ఎంత పృథక్కరణం వరకు భౌతికరాశిని కొలవడానికి వీలవుతుందో తెలియ జేస్తుంది.
2) యధార్థత దోషాల మీద ఆధారపడి ఉంటుంది.2) ఖచ్చితత్వం అనేది దోషాలపై ఆధార పడదు

ప్రశ్న 2.
కొలతలో వచ్చే వివిధ రకాల దోషాలు ఏవి?
జవాబు:
దోషాలు ప్రధానంగా మూడు రకాలు

  1. క్రమదోషాలు
  2. యాదృచ్ఛిక దోషాలు
  3. స్థూల దోషాలు

ప్రశ్న 3.
క్రమదోషాలను ఏవిధంగా కనిష్ఠం చేయవచ్చు లేదా తొలగించవచ్చు? [Mar. ’14]
జవాబు:
ప్రయోగ విధానంలో కౌశలతను పెంచుకోవడం, మంచి పరికరాలను ఎన్నుకోవడం మరియు వీలైనంత వరకు వ్యక్తిగతమైన లోపం లేకుండా చూసుకోవాలి.

ఇవ్వబడిన అమరికకు, ఈ దోషాలను అంచనావేసి, రీడింగ్లకు సరైన సర్దుబాటు తప్పనిసరిగా చేయాలి.

ప్రశ్న 4.
కొలత ఫలితాన్ని అందులో ఉండే దోషాన్ని సూచిస్తూ ఏవిధంగా నివేదిస్తారో (reported) ఉదాహరణలతో వివరించండి.
జవాబు:
ఏదైనా భౌతికరాశిని కొలిచినపుడు, ఆ రాశి యొక్క ప్రామాణిక విలువ (ప్రమాణం)తో పోల్చాలి. ప్రతి ప్రక్రియలో ఉండే దోషాలను పూర్తిగా తొలగించి కొలవడం సాధ్యం కాదు. మనం పూర్తి శ్రద్ధతో కొలిచినప్పటికీ కొలిచిన విలువ ఎల్లప్పుడూ నిజవిలువకు (లేదా) యధార్థ వేరుగా ఉంటుంది.

ప్రశ్న 5.
సార్థక సంఖ్యలంటే ఏవి? ఒక కొలత ఫలితాన్ని నివేదించేటప్పుడు అవి ఏమి సూచిస్తాయి?
జవాబు:
ఒక కొలతను సూచించే సంఖ్యలో నిశ్చయంగా తెలిసిన అంకెలు, వాటికి తోడు అదనంగా అంచనా ప్రకారం చేర్చిన అంకెలు వీటన్నింటినీ కలిపి సార్థక సంఖ్యలు (లేదా) సార్థక అంకెలు అంటారు.
ఉదాహరణ :
లఘులోలకం యొక్క ఆవర్తన కాలం 1.62, దీనిలో 1 మరియు 6 నిశ్చయంగా తెలిసిన అంకెలు, 2 అనునది అంచనా ప్రకారం చేర్చిన అంకె. కాబట్టి కొలిచిన విలువలో మూడు సార్థక సంఖ్యలు కలవు.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 6.
ప్రాథమిక ప్రమాణాలు, ఉత్పన్న ప్రమాణాల మధ్య తేడాలు రాయండి.
జవాబు:

  1. ప్రాథమిక రాశుల ప్రమాణాలను ప్రాథమిక ప్రమాణాలు అంటారు. ప్రాథమిక ప్రమాణాలను మరొక దానినుండి రాబట్టలేము. వేరొక ప్రమాణాలనుండి రాబట్టలేము.
  2. ఉత్పన్న రాశుల ప్రమాణాలను ఉత్పన్న ప్రమాణాలు అంటారు.

ప్రశ్న 7.
ఒకే భౌతికరాశికి వేరువేరు ప్రమాణాలు ఎందుకు ఉంటాయి?
జవాబు:
మనకు వేరు వేరు పద్ధతులు కలవు. అవి C.G.S పద్ధతి, M.S.K పద్ధతి, E.P.S పద్ధతి మరియు S. I పద్ధతి. అందువలన ఒకే భౌతికరాశికి వేరువేరు ప్రమాణాలు కలవు.

ప్రశ్న 8.
మితీయ విశ్లేషణ అంటే ఏమిటి?
జవాబు:
ఉత్పన్న భౌతిక రాశులను, ప్రాథమిక రాశుల ప్రమాణాలలో తెలియజేయుటను మితి విశ్లేషణ అంటారు.

మితివిశ్లేషణ ద్వారా ఇచ్చిన సమీకరణం సరైనదో, కాదో తెలుసుకోవచ్చు. ఒక పద్ధతిలో ప్రమాణాలను వేరొక పద్ధతిలోకి మార్చవచ్చు మరియు వివిధ భౌతిక రాశుల మధ్య సంబంధాన్ని సూచించే సమీకరణాలను రాబట్టవచ్చు.

ప్రశ్న 9.
కేంద్రకం వ్యాసార్ధంతో పోలిస్తే పరమాణు వ్యాసార్ధం పరిమాణ క్రమాలలో (orders of magnitude) ఎంత ఎక్కువగా ఉంటుంది?
జవాబు:
కేంద్రక వ్యాసార్థం పరిమాణం = 10-14m
పరమాణువు వ్యాసార్థం పరిమాణం = 10-10m

కాబట్టి కేంద్రకం వ్యాసార్థంతో పోలిస్తే పరమాణు వ్యాసార్థం పరిమాణక్రమాలలో 10-4 m ఎక్కువ.

ప్రశ్న 10.
ఏకీకృత పరమాణు ద్రవ్యరాశి ప్రమాణాన్ని kg లో వ్యక్తం చేయండి.
జవాబు:
ఏకీకృత పరమాణు ద్రవ్యరాశి ప్రమాణం = \(\frac{1}{12}\) × c12 – ద్రవ్యరాశి
1 a.m.u = 1.66 × 10-27 Kg

స్వల్ప సమాధాన ప్రశ్నలు

ప్రశ్న 1.
ఒక పరికరం వెర్నియర్ స్కేలు 50 విభాగాలు కలిగి ఉంది. ఇవి ప్రధాన స్కేలుపై ఉండే 49 విభాగాలతో ఏకీభవిస్తాయి. ప్రధాన స్కేలులోని ప్రతి విభాగం విలువ 0.5mm. అయితే ఈ పరికరంతో కొలిచే దూరంలో కనిష్ఠ యధార్థతారాహిత్యం (inaccuracy) ఎంత ఉంటుంది?
జవాబు:
ప్రధాన స్కేలుపై ప్రతి విభాగం విలువు = 0.5 m.m
వెర్నియర్ స్కేల్పై విభాగాల సంఖ్య = 50
వెర్నియర్ కాలిపర్స్ కనీసపు కొలత = \(\frac{S}{N}\)
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 1

కాబట్టి దూరం కొలవడంలో కనిష్ఠ యధార్ధత 0.01 m.m (లేదా)
1 M.S.D విలువ = 0.5 m.m, N = 50
N V.S.D = (N-1) M.S.D
49 M.S.D = 50 V.S.D
1 V.S.D = \(\frac{49}{50}\)M.S.D = \(\frac{49}{50}\) × 0.5 m.m

∴ కనీసపు కొలత = 1 M.S.D – 1 V.S.D = (0.5 – \(\frac{49}{50}\) × 0.5) = (1 – \(\frac{49}{50}\))0.5
= \(\frac{1}{50}\) × 0.5 = 0.01m.m
కాబట్టి దూరం, కొలవడంలో కనిష్ఠ యధార్ధత 0.01 m.m

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 2.
ప్రమాణాల ఒక వ్యవస్థలో బలానికి ప్రమాణం 100N, పొడవుకు ప్రమాణం 10m, కాలానికి ప్రమాణం 100 s. ఈ వ్యవస్థలో ద్రవ్యరాశికి ఉండే ప్రమాణం ఏది?
జవాబు:
బలం (F) = 100N; పొడవు (L) = 10m; కాలం (T) = 100s
F = ma
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 2

ప్రశ్న 3.
భూమి నుంచి ఒక గెలాక్సీ దూరం 1025 m ల క్రమంలో ఉంది. గెలాక్సీ నుంచి కాంతి మనల్ని చేరేందుకు పట్టే కాలం పరిమాణం క్రమాన్ని గణించండి.
జవాబు:
భూమి నుండి గెలాక్సీ వరకు దూరం = 1025 m
కాంతి వేగం (c) = 3 × 108m/s
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 3

ప్రశ్న 4.
భూమి-చంద్రుల మధ్య దూరం సుమారు భూవ్యాసార్థానికి 60 రెట్లు. చంద్రుడి నుంచి చూస్తే భూమి వ్యాసం సుమారుగా ఎంత ఉంటుంది ?
జవాబు:
భూమి, చంద్రుడు మధ్యదూరం
భూమి వ్యాసార్ధం = R
r = 60R = 60 × 640 × 10³ (∵ R = 6400 Km)
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 4
l = 11.16 × 10³Km ⇒ వ్యాసం( 1 ) = 11.16 × 10³Km

ప్రశ్న 5.
లోలకం 20 డోలనాలకు పట్టే కాలానికి వచ్చిన మూడు కొలతలు వరుసగా t1 = 39.6 s, t2 = 39.9 s, t3 = 39.5s. కొలతల్లోని ఖచ్చితత్వం ఎంత? కొలతల్లోని యధార్ధత ఎంత?
జవాబు:
డోలనాల సంఖ్య = 20
t1 = 39.6 Sec, t2 = 39.9 Sec, t3 = 39.5 Sec
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 5
సగటు విలువ = 39.7 Sec
ఖచ్చితత్వం = 0.1 Sec
కొలిచిన విలువ, నిజవిలువకు దగ్గరగా ఉంటే, అది యధార్థత. కాబట్టి యధార్థత 39.6 s అవుతుంది.

ప్రశ్న 6.
1 కెలోరి = 4.23, 1J = 1kg m²s-2. ద్రవ్యరాశికి ప్రమాణం a’ kg గా, పొడవుకు ప్రమాణం \(\hat{\mathbf{a}}\)m గా, కాలం ప్రమాణం \(\tilde{\mathrm{a}}\)s గా ఉండే ఒక ప్రమాణ వ్యవస్థను వాడినపుడు, కొత్త వ్యవస్థలో కెలోరికి ఉండే పరిమాణం 4.2 \(\stackrel{\wedge-1}{a} a^{\wedge}-2 \wedge 2\) అని చూపండి.
జవాబు:
1 కెలోరీ = 4.2 J ⇒ 1 J = 1 kg m²s-2
1 కెలోరీ = 4.2 kg m²s-2
క్రొత్త పద్ధతిలో, 1 కెలోరీ = 4.2 \(\hat{a} \hat{a}^2 \hat{a}^{-2}\)

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 7.
శూన్యంలో కాంతి వడి 1 ms2 అయ్యేవిధంగా పొడవుకు ఒక కొత్త ప్రమాణాన్ని ఎంచుకొన్నారు. సూర్యుడి నుంచి కాంతి భూమిని చేరేందుకు పట్టేకాలం 8 నిమిషాల 20 సెకన్లయితే కొత్త ప్రమాణాల్లో సూర్యుడు -భూమి మధ్య దూరం ఎంత?
జవాబు:
శూన్యంలో కాంతి వేగం (V) = 1m/s
పట్టుకాలం (t) = 8 నిముషాల 20 సెకండ్లు= 500 Sec
సూర్యుడు మరియు భూమి మధ్య దూరం (d) = \(\frac{V}{t}\)
d = \(\frac{1}{500}\) = 0.002m

ప్రశ్న 8.
100 ఆవర్ధనం ఉండే సూక్ష్మదర్శినిని ఉపయోగించి ఒక విద్యార్థి మానవుడి వెంట్రుక మందాన్ని కొలుస్తున్నాడు. 20 పరిశీలనల వల్ల వెంట్రుక సగటు మందాన్ని (సూక్ష్మదర్శినిలో చూసినదాని దృష్ట్వా) 3.5mm గా కనుక్కొన్నాడు. అంచనాకు వచ్చే మందం ఎంత ?
జవాబు:
సూక్ష్మదర్శిని ఆవర్ధనం (M) = 100
పరిశీలించిన మందం = 3.5 m.m
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 6

ప్రశ్న 9.
కొలవగలిగే నాలుగు రాశులు a, b, c, d లతో X అనే భౌతికరాశి కింది విధంగా సంబంధాన్ని కలిగి ఉంది.
X =a²b³c5/2d-2
a,b,c,d లను కొలవడంలో దోషశాతాలు వరుసగా 1%, 2%, 3%,4% అయితే X లో దోషశాతం ఎంత ?
జవాబు:
X =a²b³c5/2d-2

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 7

10. ఒక వస్తువు వేగం υ = At² + Bt + C అని ఇవ్వడమైంది. υ, t లను SI ప్రమాణాల్లో వ్యక్తం చేస్తే A,B,C లకు ప్రమాణాలు రాయండి.
జవాబు:
V=At² + Bt + C
సజాతీయత నియమం ప్రకారం
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 8

లెక్కలు (Problems)

ప్రశ్న 1.
P= El²m-5G-2 అనే సమాసంలో రాశులు E, l, m, G లు వరుసగా శక్తి, కోణీయ ద్రవ్యవేగం, ద్రవ్యరాశి,’ గురుత్వస్థిరాంకాలను సూచిస్తే Pఒక మితిరహిత రాశి అని చూపండి.
సాధన:
P = E L² m G-2
శక్తి (E) = [ML² T-2]
కోణీయ ద్రవ్యవేగం (L) = [ML² T-1]
ద్రవ్యరాశి = [M]
విశ్వగురుత్వ స్థిరాంకం (G) = [M-1L³ T-2]
P = [ML² T-2] [ML² T-1]² [M]-5 [M-1L³T-2]-2
P = M1 + 2 – 5 + 2 L2 + 4 – 6 T-2 – 2 + 4
P = M°L° T°
కాబట్టి P మితిరహితరాశి

ప్రశ్న 2.
కాంతి వేగం c, ప్లాంక్ స్థిరాంకం h, విశ్వ గురుత్వ స్థిరాంకం G లను ప్రాథమిక రాశులుగా తీసుకొంటే, ఈ రాశుల మితుల్లో ద్రవ్యరాశి, పొడవు, కాలాలను రాయండి. [Mar. ’13]
సాధన:
i) M ∝ GxCyhz
[M¹ L° T°] = [M-1 L³ T-2]x [LT-1]y [ML²T-1]z
[M¹ L° T°] = [M-x+z L3x+y+2z T-2x – y – z]
– x + z = 1, 3x + y + 2z = 0, – 2x – y – z = 0
ఈ సమీకరణాలను సాధించగా
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 9
M = \(\sqrt{\frac{hc}{G}}\)

ii) పొడవు (l) ∝ GxCyhz → (1)
[M°L¹T°] = [M-1 L³ T-2]x [LT-1]y [ML²T-1]z
[M°L¹T°] = M– x + z L3x+y+2zT-2x – y – z
సజాతీయత నియమం ప్రకారం
– x + z = 0, 3x + y + 2z = 0, – 2x – y – z = 0
ఈ సమీకరణాలను సాధించగా
x = \(\frac{1}{2}\), y = \(\frac{-3}{2}\), z = \(\frac{1}{2}\)
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 10

iii) కాలం (T) ∝ GxCyhz
[M°L°T¹] = [M-1 L³ T-2]x [LT-1]y [ML² T-1]z
[M°L°T¹] = M– x + z L3x + y + 2z T– 2x – y – z
సజాతీయత నియమం ప్రకారం
– x + 2 = 0, 3x + y + 2z = 0, – 2x – y – z = 1
ఈ సమీకరణాలను సాధించగా
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 11

ప్రశ్న 3.
M ద్రవ్యరాశి, R వ్యాసార్ధం కలిగి ఉండే గ్రహం చుట్టూ r వ్యాసార్ధం ఉన్న వృత్తాకార కక్ష్యలో ఒక కృత్రిమ ఉపగ్రహం పరిభ్రమిస్తుంది. మితీయ విశ్లేషణ ఆధారంగా ఉపగ్రహ కక్ష్యావర్తన కాలం T = \(\frac{\hat{e}}{R} \sqrt{\frac{r^3}{g}}\) అని చూపండి. ఇక్కడ \(\hat{\mathbf{e}}\) మితిరహిత స్థిరాంకం, g గురుత్వ త్వరణం.
సాధన:
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 12
∴ T ∝ rxGyMz
T = krxGyMz → (1)
[M°L° T¹] = K[L]x [M-1L³ T-2]y [M]z
[M°L° T¹] = K[My + z Lx + 3y T-2y]
−y + z = 0, x + 3y = 0, 1 = −2y
పై సమీకరణాలను సాధించగా
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 13

ప్రశ్న 4.
కింది సంఖ్యల్లో సార్ధక సంఖ్యలు ఎన్ని ఉన్నాయో తెలపండి.
a) 6729 b) 0.024 c) 0.08240 d) 6.032 e) 4.57 × 108
సాధన:
a) 6729 – 4 సార్థక సంఖ్యలు
b) 0.024 – 2 సార్థక సంఖ్యలు
c) 0.08240 – 4 సార్థక సంఖ్యలు
d) 6.032 – 4 సార్థక సంఖ్యలు
e) 4.57 × 108 – 3 సార్థక సంఖ్యలు

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 5.
రెండు కర్రల పొడవులు వరుసగా 12.132 cm, 12.4 cm. ఈ కర్రలను ఒకదాని చివర మరొకదాని చివరకు తాకునట్లు అమర్చితే మొత్తం పొడవు ఎంత? రెండింటిని ఒకదాని పక్క మరొకటి అమర్చితే పొడవుల్లో వ్యత్యాసం ఎంత?
సాధన:
a) కడ్డీల యొక్క పొడవులు
l1 = 12.132 cm, l2 = 12.4 cm
ఇక్కడ l2 కు దశాంశ బిందువు తర్వాత ఒక స్థానం ఉంది. కాబట్టి l1 ను రెండు దశాంశస్థానాలకు సవరించాలి.
l = l1 + l2 = 12.13 + 12.4 = 24.53
కాబట్టి ఫలితాన్ని దశాంశ బిందువు తర్వాత ఒక స్థానానికి సవరించాలి.
∴ తుది ఫలితం = 24.5 cm

b) l1 = 12.132 cm l2 = 12.4 cm
ఇక్కడ l1 ను దశాంశ బిందువు తర్వాత రెండు స్థానాలకు సవరించాలి.
l2 – l1 = 12.4 – 12.13 = 0.27
ఇప్పుడు దీనిని దశాంశ బిందువు తర్వాత ఒక స్థానానికి సవరించాలి.
l2 – l1 = 0.3

ప్రశ్న 6.
సమ ఘనం భుజం పొడవు 7.203 m. (i) ఘనం ఉపరితల వైశాల్యం, ఘనం ఘనపరిమాణాలను తగిన సార్ధక సంఖ్యలకు లెక్కించండి.
సాధన:
ఘనం యొక్క భుజం (a) = 7.203 m
(i) ఉపరితల వైశాల్యం (A) = 6a²
= 6 × (7.203)² = 311.299
7.203లో సార్ధక సంఖ్యలు నాలుగు (4), ఫలితాన్ని కూడా నాలుగు సార్ధక సంఖ్యలకు సవరించాలి.
∴ (A) = 311.3m²

(ii) ఘనం యొక్క ఘనపరిమాణం = a³
= (7.203)³ = 373.71
ఈ విలువను నాలుగు సార్థక సంఖ్యలకు సవరించాలి.
∴ V = 373.7 m³

ప్రశ్న 7.
ఒక వస్తువు ద్రవ్యరాశి 2.42 g, ఘనపరిమాణం 4.7 cm³. వాటిలోని దోషాలు వరుసగా 0.01 g, 0.1 cm³ అయితే వస్తువు సాంద్రతలో గరిష్ఠ దోషాన్ని కనుక్కోండి.
సాధన:
M = 2.42 g, V = 4.7 cm³
∆Μ = 0.01g, ∆V = 0.1 cm³
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 14

ప్రశ్న 8.
గోళం వ్యాసార్ధం కొలవడంలో దోషం 1% అయితే గోళం ఘనపరిమాణం కొలవడంలో దోషం ఎంత?
సాధన:
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 15

ప్రశ్న 9.
ద్రవ్యరాశి, వడిలో దోష శాతాలు వరుసగా 2%, 3% అయితే గతిజ శక్తిలో గరిష్ట దోషశాతం ఎంత?
సాధన:
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 16

ప్రశ్న 10.
ప్రామాణిక ఉష్ణోగ్రతా పీడనాల వద్ద ఒక మోల్ ఆదర్శవాయువు 22.4L (మోలార్ ఘనపరిమాణం) ఘనపరిమాణం ఆక్రమిస్తుంది. హైడ్రోజన్ అణు పరిమాణం సుమారుగా 1 Å అయితే హైడ్రోజన్ మోలార్ ఘనపరిమాణానికి, పరమాణు ఘన పరిమాణానికి మధ్య నిష్పత్తి ఎంత?
సాధన:
మోలార్ ఘనపరిమాణం = 22.4 లీటర్లు
= 22.4 × 1000 c.c.
= 22.4 × 10-3
హైడ్రోజన్ అణువు వ్యాసము
= 1A ° 10-10 m
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 17

పరమాణు ఘనపరిమాణం = V × అవగాడ్రో సంఖ్య (N)
= 0.5233 × 10-30 × 6.023 × 1023 = 3.151 × 10-7

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 18

అదనపు లెక్కలు (Additional Problems)

ప్రశ్న 1.
ఖాళీలను పూరించండి.
a) 1 cm భుజం పొడవు ఉండే సమఘనం m ఘనపరిమాణం ……….. m³
b) 2.0 cm వ్యాసార్ధం, 10.0 cm ఎత్తు ఉండే ఘన స్తూపం ఉపరితల వైశాల్యం (mm)²
c) 18 km h-1 వడితో చలిస్తున్న వాహనం 1 s లో ప్రయాణించే దూరం…… m
d) సీసం సాపేక్ష సాంద్రత 11.3 అయితే దాని సాంద్రత …………… g cm³ లేదా …………… kg m-3
సాధన:
a) పొడవు L = 1 cm = 10-2 m
ఘనం ఘనపరిమాణం L³ = (10-2
= 10-6

b) r = 2 cm = 20 m.m
h = 10.0 cm = 100 m.m
ఘనస్థూపం ఉపరితల వైశాల్యం 2πr × h
= 2 × \(\frac{22}{7}\) × 20 × 100
= 1.26 × 104 m.m²

c) వడి (V) = 18 kmph
= \(\frac{18\times1000}{60\times60}\) = 5 m/s
∴ 1 సెకనులో ప్రయాణించిన దూరం = 5 m

d) సాపేక్ష సాంద్రత = 11.3 gm/c.c
= \(\frac{11.3\times10^{3}}{(10^{-2})^3}\)
= 11.3 × 10³ kg/m³

ప్రశ్న 2.
ప్రమాణాలను తగురీతిలో పరివర్తన చేయడం ద్వారా ఖాళీలను పూరించండి.
a) 1 kg m² s-2 = …. 9 cm² 5-2
b) 1 m = …. 1y (కాంతి సంవత్సరాలు)
c) 3.0 ms-2 = …. km h-2
d) G = 6.67 × 10-11 N m² (kg)-2 (cm)³ s-2 g-1.
సాధన:
a) 1 kg m² s-2
= 1 × 10³ × (10²)² s-2 = 107 gm cm² s-2

b) 1 కాంతి సంవత్సరం = 9.46 × 1015 m
∴ 1m = \(\frac{1}{9.46\times10^{-16}}\) కాంతి సంవత్సరం
= 1.053 × 10-16 కాంతి సంవత్సరం

c) 3ms-2 = 3 × 10-3 km(\(\frac{1}{60\times60}\)h)-2
= 3 × 10-3 × 3600 × 3600 km h-2
= 3.888 × 104 km h-2

d) G = 6.67 × 10-11 Nm² kg-2
= 6.67 × 10-11 (kg ms-2) m² kg-2
= 6.67 × 10-11 m³ s-2 kg-1
= 6.67 × 10-11 (100 cm)³ s-2 (1000 g)-1
= 6.67 × 10-8 cm³ s-2 g-1

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 3.
శక్తి లేదా ఉష్ణానికి ప్రమాణం కెలోరి. దీని విలువ సుమారు 4.2 J ఇక్కడ 1J = 1 kg m²s-2. ద్రవ్యరాశికి ప్రమాణం a kg, పొడవుకు ప్రమాణం b m, కాలానికి ప్రమాణం g s. అయ్యే ప్రమాణాల వ్యవస్థను వాడినామనుకోండి అప్పుడు కొత్త ప్రమాణాల పదాలలో కెలోరికి ఉండే పరిమాణం 4.2 a-1b-2 g² అని చూపండి.
సాధన:
1 కెలోరీ = 4.2 J = 4.2 Kg m²S-2
ద్రవ్యరాశి క్రొత్త ప్రమాణం a kg
∴ 1 kg = \(\frac{1}{a}\) = a-1
అదేవిధంగా 1m = b-1 ద్రవ్యరాశి క్రొత్త ప్రమాణం
1s = g-1 కాలం క్రొత్త ప్రమాణం
1 కెలోరీ = 4.2 (a-1) (b-1)² (g-1)-2
1 కెలోరీ = 4.2a-1 b-2g², శక్తి యొక్క క్రొత్త ప్రమాణం

ప్రశ్న 4.
కింది ప్రవచనాన్ని (statement) స్పష్టంగా వివరించండి :
” పోలికకు అవసరమయ్యే ప్రామాణికాన్ని నిర్దేశించకుండా మితీయరాశిని పెద్దది లేదా చిన్నది అని పిలవడం అర్థరహితం” దీన్ని దృష్టిలో ఉంచుకొని కింది ప్రవచనాలను అవసరమైన చోట సరిచేసి తిరిగి రాయండి.
a) పరమాణువులు అతిచిన్న వస్తువులు.
b) జెట్ విమానం ఎక్కువ వడితో చలిస్తుంది.
c) బృహస్పతి ద్రవ్యరాశి చాలా ఎక్కువ.
d) ఈ గదిలోని గాలి అధిక సంఖ్యలో అణువులను కలిగి ఉంది.
e) ఎలక్ట్రాన్ కంటే ప్రోటాన్ ద్రవ్యరాశి చాలా ఎక్కువ.
f) కాంతి వేగం కంటే ధ్వని వేగం చాలా తక్కువ.
సాధన:
i) పై నిర్వచనం సరియైనది. మితిరహితరాశులను కొన్ని ప్రమాణాలకు లోబడి పెద్దవా (లేదా) చిన్నవా అని పోల్చుతారు.
ఉదా : కోణం మితిరహితరాశి, ∠θ = 60° అనునది ∠θ = 30° కన్నా ఎక్కువ కాని ∠θ = 90° కన్నా తక్కువ.

ii) a) పరమాణువు పరిమాణం, ఒక గుండు సూదిమొన కన్నా చాలా చిన్నది.
b) సూపర్ ఫాస్ట్ రైలు కన్నా జెట్ విమానం చాలా వేగంగా పోతుంది.
c) భూమి ద్రవ్యరాశి కన్నా అంగారకుడి ద్రవ్యరాశి బాగా ఎక్కువ.
d) ఒక మోల్ వాయువులోని అణువుల కన్నా, గదిలో
గాలిలో అణువుల సంఖ్య ఎక్కువ.
e) పై స్టేట్మెంట్ సరియైనది.
f) పై స్టేట్మెంట్ సరియైనది.

ప్రశ్న 5.
శూన్యంలో కాంతి వడి 1ms-2 అయ్యేవిధంగా పొడవుకు ఒక కొత్త ప్రమాణాన్ని ఎంచుకొన్నారు. సూర్యుడి నుంచి కాంతి భూమిని చేరేందుకు పట్టేకాలం 8 నిమిషాల 20 సెకన్లయితే కొత్త ప్రమాణాల్లో సూర్యుడు-భూమి మధ్య దూరం ఎంత?
సాధన:
శూన్యంలో కాంతివేగం c = 1 పొడవు క్రొత్త ప్రమాణం s-1
కాంతి సూర్యుడి నుండి భూమిని చేరుటకు పట్టుకాలం
t = 8 నిమిషాల 20 సెకండ్లు = 8 × 60 + 20
= 500 సెకండ్ల
భూమి, సూర్యుడి మధ్యదూరం (x) = C × t
∴ పొడవు క్రొత్త ప్రమాణం s-1 × 500
= 500 పొడవు క్రొత్త ప్రమాణం

ప్రశ్న 6.
పొడవును కొలవడానికి కింది వాటిలో ఏది చాలా ఖచ్చితమైన పరికరం?
a) కదిలే స్కేలుపై 20 వెర్నియర్ విభాగాలు ఉండే వెర్నియర్ కాలిపర్స్
b) 1 mm పిచ్, 100 తలస్కేలు విభాగాలు ఉండే స్కూృగేజి.
c) కాంతి తరంగదైర్ఘ్యవిలువకు తక్కువ/సమానం వరకు పొడవును కొలిచే దృక్ సాధనం.
సాధన:
a) వెర్నియర్ కాలిపర్స్ కనీసపు కొలత
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 19
b) స్క్రూగేజి కనీసపు కొలత
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 20
c) కాంతి సంవత్సరం
λ = 10-5 cm = 0.00001 cm
ఎక్కువ యదార్థ కలది దృశావస్తువు

ప్రశ్న 7.
100 ఆవర్ధనం ఉండే సూక్ష్మదర్శిని ఉపయోగించి ఒక విద్యార్థి మానవుడి వెంట్రుక మందాన్ని కొలుస్తున్నాడు. 20 పరిశీలనల వల్ల వెంట్రుక సగటు మందాన్ని (సూక్ష్మదర్శినిలో చూసినదాని దృష్ట్యా) 3.5 mm.గా కనుక్కొన్నాడు. అంచనాకు వచ్చే మందం ఎంత
సాధన:
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 21

ప్రశ్న 8.
కింది వాటికి సమాధానం రాయండి:
a) నీకు ఒక దారం, మీటరు స్కేలును ఇస్తే దారం వ్యాసాన్ని ఏవిధంగా అంచనా వేస్తావు?
b) ఒక స్కూృగేజి పిచ్ 1.0 mm, వృత్తాకార స్కేలుపై విభాగాలు 200. వృత్తాకార స్కేలుపై విభాగాల సంఖ్యను అనియతంగా పెంచడం ద్వారా స్కూృగేజి యధార్ధతను పెంచడం సాధ్యమని నీవు అనుకొంటున్నావా?
c) వెర్నియర్ కాలిపర్స్ సహాయంతో పలుచని ఇత్తడి కడ్డీ సగటు వ్యాసాన్ని నిర్ణయించవలసి ఉంది. 5 కొలతల సమితి కంటే 100 కొలతల సమితితో వచ్చే అంచనా విలువ ఎక్కువ నమ్మదగినదని మనమెందుకు ఆశిస్తాం?
సాధన:
a) ద్వారం వ్యాసం చాలా చిన్నది కాబట్టి దానిని మీటరు స్కేలు ఉపయోగించి కొలవలేము. మీటరు స్కేలుపై దారాన్ని అనేకచుట్లుగా చుట్టాలి. ఆ చుట్లు బాగా దగ్గర దగ్గరగా చుట్టాలి.
n చుట్లు గల దారం పొడవు l
∴ దారం వ్యాసం = \(\frac{1}{n}\)

b) కనీసపు కొలత
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 22
వృత్తాకార స్కేలుపై విభాగాల సంఖ్య పెరిగితే, కనీసపు కొలత తగ్గుతుంది. కాబట్టి ఖచ్చితత్వం పెరుగు తుంది. కాని ప్రాయోగికంగా సాధారణ మానవుని కంటితో రీడింగ్లను ఖచ్చితంగా కొలవలేము.

c) తక్కువ పరిశీలనలు (5) కన్నా ఎక్కువ పరిశీలనలు (100) ఆధారపడదగినవిగా ఉంటాయి. దీనికి కారణం ధనయాదృచ్ఛిక దోషం, రుణయాదృచ్ఛిక దోషం ఒకే పరిమాణంలో ఉంటాయి. కాబట్టి పరిశీలనల సంఖ్య పెరిగే కొద్దీ యాదృచ్ఛిక దోషాలు తగ్గి, విలువలు ఆధారపడదగినవిగా ఉంటాయి.

ప్రశ్న 9.
35 mm స్లైడుపై ఒక ఇంటి ఛాయాచిత్రం వైశాల్యం 1.75 cm². ఆ స్లెడును తెరపై ప్రొజెక్ట్ చేసినపుడు ఇంటి వైశాల్యం 1.55 m² గా ఉంది. ప్రొజెక్టర్-తెర అమరిక రేఖీయ ఆవర్ధనం ఎంత?
సాధన:
వస్తువు వైశాల్యం = 1.75 cm²
ప్రతిబింబం వైశాల్యం = 1.55 m² = 1.55 × 104cm²
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 23

ప్రశ్న 10.
కింద ఇచ్చిన వాటిలో సార్ధక సంఖ్యలు ఎన్ని ఉన్నాయి?
a) 0.007 m², b) 2.64 ×1024 kg, c) 0.2370 g cm-3, d) 6.320 J, e) 6.032 N m-2, f) 0.0006032 m²
సాధన:
ఇచ్చిన సంఖ్యలకు ఈ క్రింది సార్ధక సంఖ్యలు కలవు.
a) ఒకటి b) మూడు c) నాలుగు d) నాలుగు e) నాలుగు f) నాలుగు.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 11.
దీర్ఘచతురస్రాకార లోహ పలక పొడవు, వెడల్పు, మందాలు వరుసగా 4.234m, 1.005 m, 2.01 cm లు సరయిన సార్ధక సంఖ్యల వరకు ఆ పలక వైశాల్యం, ఘనపరిమాణాలను రాయండి. సాధన:
ఇక్కడ పొడవు l = 4.234 m
వెడల్పు b = 1.005 m
మందం t = 2.01 cm = 2.01 × 10-2 m
పలక వైశాల్యం = 2 (l × b + b × t + t × l)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.021 × 4.234)
=2(4.3604739)
= 8.7209478 m²

వైశాల్యం మూడు సార్ధక సంఖ్యలు కలిగి ఉండాలి. కావున వీటిని సవరించాలి.
వైశాల్యం = 8.72 m²
ఘనపరిమాణం (V) = l × b × t
V = 4.234 × 1.005 × 0.0201
= 0.0855289
= 0.855 m³
దీనిలో మూడు సార్ధక సంఖ్యలు కలవు.

ప్రశ్న 12.
ఒక పెట్టెను కిరాణా షాపుదారు వాడే త్రాసుతో తూస్తే వచ్చిన ద్రవ్యరాశి 2.300 kg. ఇప్పుడు ఈ పెట్టెకు 20.15 g, 20.17 g ద్రవ్యరాశులు గల రెండు బంగారు ముక్కలను కలిపారు. (a) పెట్టె మొత్తం ద్రవ్యరాశి, (b) ముక్కల ద్రవ్యరాశుల్లో వ్యత్యాసాన్ని సరయిన సార్ధక సంఖ్యల వరకు రాయండి.
సాధన:
పెట్టె ద్రవ్యరాశి m = 2.3 kg
ఒక బంగారం ముక్క ద్రవ్యరాశి
m1 = 20.15 g = 0.02015 kg
మిగిలిన బంగారం ద్రవ్యరాశి
m2 = 20.17g = 0.02017 kg

a) మొత్తం ద్రవ్యరాశి = m + m1 + m2
= 2.3 +0.02015 +0.02017
= 2.34032 kg
ఫలితాన్ని దశాంశ స్థానం తర్వాత ఒక సంఖ్యకు సవరించాలి.
మొత్తం ద్రవ్యరాశి = 2.3 kg

b) ద్రవ్యరాశులలో తేడా
= m2 – m1 = 20.17 – 20.15
= 0.02gm

ప్రశ్న 13.
P అనే భౌతికరాశి a, b, c, d అనే నాలుగు పరిశీలించగలిగే రాశులతో కిందివిధమైన సంబంధాన్ని కలిగి ఉంది :
P = a³b²/(√cd)
a,b,c,d లకొలతల్లోని దోషశాతాలు వరుసగా 1%, 3%, 4%, 2% అయితే P లోని దోషశాతం ఎంత ? పై సంబంధం ఉపయోగించి లెక్కించిన P విలువ 3.763 అయితే, ఫలితాన్ని నీవు ఏ విలువ వరకు సవరిస్తావు ?
సాధన:
ఇక్కడ P = \(\frac{a^3 b^2}{\sqrt{c} d}\)
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 24

ఫలితం Pలో రెండు సార్ధక సంఖ్యలు కలవు కాబట్టి 3.763ని సవరిస్తే 3.8 అవుతుంది.

ప్రశ్న 14.
ముద్రణా దోషాలు అనేకంగా ఉండే పుస్తకంలో ఒక నిర్దిష్ట ఆవర్తన చలనం చేస్తున్న కణం స్థానభ్రంశానికి నాలుగు భిన్న ఫార్ములాలు ఉన్నాయి. అవి
a) y = a sin 2 p t/T, b) y = a sin vt, c) y = (a/T) sin t/a, d) y = (a√2) (sin 2 πt /T + cos 2 πt /T)
(a = కణం పొందే గరిష్ఠ స్థానభ్రంశం, v = కణం వడి, T = ఆవర్తన కాలం) మితుల దృష్ట్యా తప్పు అయిన ఫార్ములాలను కొట్టి వేయండి.
సాధన:
త్రికోణమితి ప్రమేయం మితిరహితరాశి అనగా కోణం మితిరహిత రాశి
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 25
కాబట్టి (ii) మరియు (iii) సూత్రాలు తప్పు.

ప్రశ్న 15.
భౌతికశాస్త్రంలో ఒక ప్రసిద్ధమైన సంబంధం కణ ‘విరామ ద్రవ్యరాశి’ m0, చలిస్తున్నప్పుడు ద్రవ్యరాశి m లను కణ వడి v కాంతి వడి c పదాలతో కలుపుతుంది. (ఈ సంబంధం ఆల్బర్ట్ ఐన్స్టీన్ ప్రతిపాదించిన ప్రత్యేక సాపేక్షతా సిద్ధాంతం పర్యవసానంగా మొట్టమొదట వెలువడింది) ఒక బాలుడు ఈ సంబంధాన్ని దాదాపు సరిగానే జ్ఞప్తికి తెచ్చుకొన్నాడు కానీ స్థిరాంకం c ని ఎక్కడ ఉంచాలో మరచిపోయి కింది విధంగా రాశాడు.
m = \(\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\)
ఇందులో కనిపించకుండా పోయిన ఁని ఎక్కడ ఉంచాలో ఊహించండి.
సాధన:
సజాతీయత సూత్రం ప్రకారం, MLT యొక్క ఘాతాలు
ఇరువైపులా సమానం. R.H.S వైపు లవంలో (1 – v²)1/2

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 26

ప్రశ్న 16.
పరమాణు మానంపై (atomic scale) అనువుగా ఉండే పొడవు ప్రమాణం ఆంగ్జామ్, దీన్ని AC తో సూచిస్తారు. 14° = 10-10 m. హైడ్రోజన్ పరమాణు పరిమాణం సుమారు 0.5A° . ఒక మోల్ హైడ్రోజన్ పరమాణువుల మొత&ం పరమాణు ఘనపరిమాణం m³ ల్లో ఎంత ఉంటుంది?
సాధన:
ఇక్కడ r = 0.5 A° = 0.5 × 10-10 m
ప్రతి హైడ్రోజన్ పరమాణువు ఘనపరిమాణం = \(\frac{4}{3}\)πr³
\(\frac{4}{3}\) × 3.14 × (0.5 × 10-10)³ = 5.236 × 10-31 3 m³
ఒక గ్రామ్ మోల్ హైడ్రోజన్ పరమాణువుల సంఖ్య అవగాడ్రో సంఖ్య = 6.023 × 1023

ఒక గ్రామ్ – మోల్ హైడ్రోజన్ పరమాణువు ఘన పరిమాణం
= 5.236 × 10-31 × 6.023 × 1023
= 3.154 × 10-7

ప్రశ్న 17.
ప్రామాణిక ఉష్ణోగ్రతా పీడనాల వద్ద ఒక మోల్ ఆదర్శవాయువు 22.4L (మోలార్ ఘనపరిమాణం) ఆక్రమిస్తుంది. హైడ్రోజన్ అణు పరిమాణం సుమారుగా 1A° అయితే హైడ్రోజన్ మోలార్ ఘనపరిమాణానికి, పరమాణు ఘనపరిమాణానికి మధ్య నిష్పత్తి ఎంత? ఈ నిష్పత్తి అంత పెద్దదిగా ఎందుకు ఉంది?
సాధన:
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 27
ఈ నిష్పత్తి పరమాణువుల అంతర దూరాల కన్నా అధికం.

ప్రశ్న 18.
సాధారణంగా మనం పరిశీలించే ఈ అంశాన్ని స్పష్టంగా వివరించండి. వేగంగా ప్రయాణిస్తున్న రైలు కిటికీలో నుంచి బయటికి నీవు చూస్తే, దగ్గరలో ఉండే చెట్లు, ఇళ్ళు మొదలైనవి రైలు ప్రయాణ దిశకు వ్యతిరేక దిశలో వేగంగా కదులుతున్నట్లుగా అగుపిస్తాయి. కానీ, దూరంగా ఉండే వస్తువులు (కొండ శిఖరాలు, చంద్రుడు నక్షత్రాలు మొదలైనవి) స్థిరంగా ఉన్నట్లుగా అగుపిస్తాయి. (నిజానికి, నీవు చలిస్తున్నావని నీకు తెలుసు కాబట్టి ఈ దూర వస్తువులు నీతోపాటే చలిస్తున్నట్లు అనిపిస్తుంది)
సాధన:
వస్తువుకు, నేత్రానికి గీసే సరళరేఖను దృష్టిరేఖ అంటారు. రైలు వేగంగా కదిలితే, దగ్గరలో ఉన్న చెట్లపై దృష్టిరేఖ మారును. అందుకని చెట్లు వెనక్కి పోవుచున్నట్లు కనిపిస్తుంది.

దూరపు వస్తువులపై (కొండలు, చంద్రుడు, నక్షత్రాలు) దృష్టిరేఖ మారదు. కాబట్టివాటి దిశ మారదు. అందువలన పర్వతాలు, చంద్రుడు, నక్షత్రాలు వంటి దూరంగా ఉన్నవి నిశ్చలంగా ఉంటాయి.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 19.
విభాగం 2.3.1 లో చర్చించిన దృష్టి విక్షేప సూత్రం సుదూరంగా ఉండే నక్షత్రాల దూరాలను నిర్ణయించడానికి ఉపయోగించాం. 6 నెలల ఎడంలో సూర్యుడి చుట్టూ ఉండే దాని కక్ష్యలో గల భూమి రెండు స్థానాలను కలిపే రేఖ ఆధారరేఖ (baseline) AB. అంటే ఆధారరేఖ సుమారు భూకక్ష్య వ్యాసానికి సమానం (≈3 × 1011 m). అయితే మనకు దగ్గర్లో ఉండే నక్షత్రాలు కూడా ఎంత దూరంలో ఉంటాయంటే అంతటి పొడవైన ఆధారరేఖతో అవి కలగచేసే ‘దృష్టి విక్షేపం’ చాపంలో 1″ (సెకను) క్రమంలో మాత్రమే ఉంటుంది. ఖగోళదూరాలకు ‘పార్సెక్’ అనేది అనుకూలమైన ‘పొడవు ప్రమాణం’. ఇది భూమి నుంచి సూర్యుడి వరకు గల దూరానికి సమానంగా ఉండే ఆధారరేఖ యొక్క రెండు వ్యతిరేక చివరలతో చాపపు 1″ (సెకను) పారలాక్స్ (కోణం) (దృష్టి విక్షేపం) ను చూపగలిగే వస్తువు దూరం. ఒక పార్సెక్ విలువ మీటర్లలో ఎంత?
సాధన:
ఆధారరేఖ పొడవు = భూమి నుండి సూర్యుడి వరకు పొడవు
= 1 A.U = 1.5 × 1011m
దృష్టికోణం θ = 111
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 28
కాబట్టి 1 పార్సెక్ 3.1 × 1016 m

ప్రశ్న 20.
మన సౌరవ్యవస్థకు అత్యంత సమీపంలో ఉండే నక్షత్రం దూరం 4.29 కాంతి సంవత్సరాలు. ఈ దూరం ‘పార్సెక్’ లలో ఎంత? ఆల్పాసెంటారీ (Alpha Centauri) అని పిలువబడే నక్షత్రం సూర్యుని చుట్టూ ఉండే కక్ష్యలో భూమి 6 నెలల ఎడంలో ఉన్నప్పుడు ఉండే రెండు స్థానాల నుంచి చూసినప్పుడు ఎంత దృష్టి విక్షేపాన్ని కలగచేస్తుంది?
సాధన:
x = 4.29 కాంతి సంవత్సరాలు
= 4.29 × 9.46 × 1015 m
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 29
= 1.512 సెకన్లు

ప్రశ్న 21.
విజ్ఞానశాస్త్రానికి భౌతికరాశుల కచ్చిత కొలతలు ఆవశ్యం. ఉదాహరణకు ఒక విమాన వడిని నిశ్చయంగా తెలుసుకోవాలంటే స్వల్ప కాల వ్యవధుల ఎడంలో విడిపోయి ఉన్న దాని స్థానాలను కనుక్కోవడానికి ఒక యధార్థ పద్ధతి మనకు ఉండాలి. రెండో ప్రపంచ యుద్ధంలో రాడారును ఆవిష్కరించడం వెనుక ఉన్న నిజమైన ఉద్దేశం ఇదే. ఆధునిక విజ్ఞానశాస్త్రంలో ఎక్కడైతే పొడవు, కాలం, ద్రవ్యరాశి మొదలైనవాటి కచ్చిత కొలతలు అవసరమవుతాయో అలాంటి భిన్న ఉదాహరణల గురించి ఆలోచించండి. దాంతో పాటు అక్కడ అవసరమయ్యే కచ్చితత్వం పరిమాణాత్మక భావనను ఇవ్వగలిగిన చోట ఇవ్వండి.
సాధన:
సైన్స్ మరియు భౌతికశాస్త్ర నియమాలు అభివృద్ధి చెందాలంటే పొడవు, ద్రవ్యరాశి మరియు కాలాలను ఖచ్చితంగా కొలవాలి.

ఉదాహరణకు లేసర్ కిరణంతో భూమి నుండి, చంద్రుడు వరకు దూరాన్ని కొలిచామనుకోండి. అందుకు కాలంను ఖచ్చితంగా కొలవాలి. అదేవిధంగా దూరం, విమాన వేగాన్ని రాడార్తో కొలవడం మొదలగువాటికి కాలంను ఖచ్చితంగా కొలవాలి. నక్షత్రాల మధ్య దూరాలను కొలుచుటకు దృష్టికోణంను ఖచ్చితంగా కొలవాలి.

స్ఫటికాలలో పరమాణువులలో అంతర దూరాలు ఖచ్చితంగా కొలవాలి. ద్రవ్యరాశి స్పెక్ట్రోమీటరు ఉపయోగించి పరమాణువుల ద్రవ్యరాశిని ఖచ్చితంగా కొలవవచ్చు.

ప్రశ్న 22.
మూల భావనలను, సాధారణ పరిశీలనలను ఉపయోగిస్తూ, రాశులకు ఉజ్జాయింపు అంచనాలను కట్టగలగడం కూడా అంతే ముఖ్యం. ఈ కింది వాటిని అంచనావేసే మార్గాలను ఆలోచించండి. (ఎక్కడైతే అంచనా పొందడం కష్టమో, అక్కడ రాశి యొక్క విలువలో పైహద్దు (upper bound) ను పొందడానికి ప్రయత్నించండి)
a) రుతుపవనాల కాలంలో భారతదేశంపై ఆవరించి ఉండే వర్షాన్ని తీసుకొని రాగలిగే మేఘాల మొత్తం ద్రవ్యరాశి.
b) ఏనుగు ద్రవ్యరాశి
c) తుఫాన్ సమయంలో పవన వడి
d) నీ తలపై ఉండే వెంట్రుకల సంఖ్య
e) మీ తరగతి గదిలోని గాలి అణువుల సంఖ్య.
సాధన:
a) రుతుపవనాల కాలంలో, మెటీరియాలజిస్ట్ 100 cm వర్షపాతం కురిసినట్లుగా నమోదు చేశాడు.
= 100 cm = 1m.
దేశం వైశాల్యం,
A = 3.3 మిలియన్ × km²
= 3.3 × 106 (10³)²
= 3.3 × 1012
వర్షపు నీరు ఘనపరిమాణం
v = Axh = 3.3 × 1012 × 1m³
నీటిసాంద్రత ρ = 10 km/m³
వాన నీటి ద్రవ్యరాశి
= vρ = 3.3 × 1012 × 10³
= 3.3 × 1015 kg
ఇది భారతదేశం అంతటా కురిసిన వర్షం మొత్తం ద్రవ్యరాశి.

b) ఏనుగు ద్రవ్యరాశి కొలవాలంటే, వైశాల్యం (A) తెలిసిన పడవను తీసుకోవాలి. నీటిలో పడవ లోతును తెలుసుకోవాలి. అది x1 అనుకొనుము. పడవ వలన స్థానభ్రంశం చెందిన నీటి ఘన పరిమాణం V1 = Ax1. ఇప్పుడు పడవలోకి ఏనుగును ఎక్కించాలి. పడవ నీటిలోకి దిగుతుంది. ఇప్పుడు పడవ నీటిలోకి ఎంత లోతు ఉన్నది కొలవాలి. అది x2 గా తీసుకోవాలి.

పడవ మరియు ఏనుగు వలన స్థానభ్రంశం చెందిన నీటి యొక్క ఘనపరిమాణం V2 = Ax2

ఏనుగువల్ల స్థానభ్రంశం చెందిన నీటి ఘన పరిమాణం (V) = V2 – V1 = A(x2 – x1)

నీటిసాంద్రత p, అయిన ఏనుగు ద్రవ్యరాశి
= తొలగిన నీటి ద్రవ్యరాశి
= Vρ = A(x2 – x1

c) బెలూన్లో వాయువును నింపి గాలి వేగాన్ని తుఫాన్ సమయంలో కొలవవచ్చు. OA అనునది సాధారణ స్థితిలో వాయుబెలూన్ ఉంది. గాలి కుడివైపుకు వీస్తే ఒక సెకన్లో B వద్దకు బెలూన్ చేరినది. అప్పుడు ∠AOB = θను కొలవాలి. బెలూన్ ఎత్తు h అయిన AB = d = hθ. ఇది బెలూన్ ఒక సెకనులో ప్రయాణించిన దూరం. ఇదే గాలివేగం.
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 30

d) ముందుగా తల వైశాల్యాన్ని కొలవాలి. అది A అనుకొనుము. స్క్రూగేజిని ఉపయోగించి, వెంట్రుక మందాన్ని కొలవాలి (d).
వెంట్రుక అడ్డుకోత వైశాల్యం = πd²
తలపై ఉన్న వెంట్రుకల సంఖ్య
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 31
కాబట్టి లెక్కల ప్రకారం తలపై సుమారుగా ఒక మిలియన్ వెంట్రుకలు ఉంటాయి.

e) ముందుగా గది ఘనపరిమాణంను కొలవాలి. N.T.P వద్ద ఒక మోల్ వాయువు ఘనపరిమాణం 22.4 లీటర్లు అనగా 22.4 × 10³ m³
∴ వాయు అణువుల సంఖ్య
= 22.4 × 10-3 m³ = 6.023 × 1023
V ఘనపరిమాణం గల తరగతి గదిలో వాయు
అణువుల సంఖ్య = \(\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}} \times v\)

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 23.
అతి వేడిగా ఉండే ప్లాస్మా (అయనీకృత పదార్థం) స్థితిలో సూర్యుడు ఉంటాడు. సూర్యుడి అంతర్భాగ ఉష్ణోగ్రత 107 K కు మించి ఉంటుంది. బాహ్య ఉపరితల ఉష్ణోగ్రత సుమారు 6000 K. ఇంతటి అధిక ఉష్ణోగ్రతల వద్ద ఏ పదార్థం కూడా ఘన లేదా ద్రవ ప్రావస్థలో ఉండదు. ఘనపదార్థాలకు, ద్రవాలకు లేదా వాయువులకు ఉండే సాంద్రత వ్యాప్తిలో సూర్యుని ద్రవ్యరాశి సాంద్రత ఏ వ్యాప్తిలో ఉంటుందని లేదా ఉండవచ్చని మీరు భావిస్తున్నారు. మీరు ఊహించిన విలువ సరియైందో కాదో కింది దత్తాంశంతో సరిచూడండి. సూర్యుని ద్రవ్యరాశి : 2.0 × 1030 kg, సూర్యుని వ్యాసార్ధం = 7.0 × 108 m.
సాధన:
m = 2.0 × 1030 kg, R = 7.0 × 108
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 32
ఇది ఘన మరియు ద్రవాల సాంద్రత. వాయువులకు కాదు. సూర్యుడిపై బాహ్యపొరలపై లోపలివైపుకు గురుత్వాకర్షణ బలాలు ఉండుటచే సూర్యుడిలో అధిక సాంద్రత ఉంటుంది.

ప్రశ్న 24.
బృహస్పతి గ్రహం భూమి నుంచి 824.7 మిలియన్ కిలోమీటర్ల దూరంలో ఉన్నప్పుడు కోణీయ వ్యాసాన్ని కొలిస్తే 35.72″, గా వచ్చింది. బృహస్పతి వ్యాసాన్ని లెక్కించండి.
సాధన:
ఇక్కడ r = 824.7 × 106 km,
θ = 35.72
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 33

ప్రశ్న 25.
వర్షంలో వడితో వడివడిగా నడిచే వ్యక్తి తన గొడుగును నిలువు (క్షితిజ లంబం) తో q కోణం చేసే విధంగా ముందుకు వంచాలి. q, υ ల మధ్య సంబంధాన్ని ఒక విద్యార్థి tan q = υ గా ఉత్పాదించి ఈ సంబంధం సరియైన అవధి కలిగి ఉందని సరిచూశాడు. ఈ అవధి ఆశించినట్లుగానే ఉందని కనుక్కొంటాడు. అవధులు υ > 0, q > 0 (ఎలాంటి ప్రబలమైన గాలి వీచడం లేదని వర్షం నిశ్చలంగా ఉండే వ్యక్తి దృష్ట్యా నిలువుగానే పడుతుందని మనం భావించుకుంటున్నాం) ఈ సంబంధం సరియైందేనని మీరు భావిస్తున్నారా ? కాకపోయి నట్లయితే, సరియైన సంబంధాన్ని ఊహించండి.
సాధన:
tan θ = υ,
υ → 0; θ → 0.
R.H.S = tan θ = [M°L°T°] మరియు
L.H.S = υ = [M° L¹T-1]
మితులపరంగా ఈ సంబంధం సరియైనది కాదు. సరియైన సంబంధం
tan θ = \(\frac{v^2}{\mathrm{rg}}\)

ప్రశ్న 26.
రెండు సీజియం గడియారాలను నిరాటంకంగా 100 సంవత్సరాలు నడిపిస్తే, వాటి కాలాల్లో వచ్చే మార్పు కేవలం 0.025 అని నొక్కిచెప్పారు. 1 s కాలవ్యవధిని కొలవడంలో ప్రామాణిక సీజియం గడియారం యదార్థతకు ఇది ఏ అర్ధాన్ని ఇస్తుంది?
సాధన:
100 సంవత్సరాలకు దోషం = 0.02s
1 సెకనులో దోషం
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 34
కాబట్టి ప్రామాణిక సీజియమ్ గడియారంలో కొలచే కాలం వ్యవధి 1 సెకనుకు యదార్థత 10-12 s.

ప్రశ్న 27.
సోడియం పరమాణువు పరిమాణాన్ని 2.5 A° గా పరిగణిస్తూ, సోడియం పరమాణువు సగటు ద్రవ్యరాశి సాంద్రతను అంచనా వేయండి. (అవొగాడ్రో) సంఖ్య, సోడియం పరమాణు ద్రవ్యరాశులకు తెలిసిన విలువలను ఉపయోగించండి.) ఈ సాంద్రత విలువను స్ఫటిక ప్రావస్థలోని సోడియం సాంద్రత విలువ 970 kg m-3 తో పోల్చండి. ఈ రెండు సాంద్రతలు ఒకే పరిమాణ క్రమాన్ని కలిగి ఉన్నాయా? ఉంటే, ఎందుకు?
సాధన:
పరమాణు ఘనపరిమాణం = \(\frac{4}{3}\) πR³ × N
= \(\frac{4}{3}\times\frac{22}{7}\) (1.25 × 10-10)³ × 6.023 × 1023
4.93 × 10-6
సగటు ద్రవ్యరాశి సాంద్రత
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 35
= 4.67 × 103 kg/m³
ఈ రెండు సాంద్రతలు ఒకే క్రమంలో లేవు. ఇది స్ఫటికాలలో పరమాణువుల అంతర దూరాల దశను తెలుపును.

ప్రశ్న 28.
కేంద్రక మానం (nuclear scale) లో అనుకూలమైన పొడవు ప్రమాణం ఫెర్మీ. 1f = 10-15 m. కేంద్రక పరిమాణాలు ఉజ్జాయింపుగా అనుసరించే అనుభవిక సంబంధం : r = roA1/3
ఇక్కడ r కేంద్రక వ్యాసార్ధం, దాని ద్రవ్యరాశి, సంఖ్య ro అనేది 1.2 f కు దాదాపు సమానమైన స్థిరాంకం. ఈ నియమం ఆధారంగా భిన్న కేంద్రకాలకు ద్రవ్యరాశి సాంద్రత దాదాపు స్థిరాంకం అని చూపండి. సోడియం కేంద్రకం ద్రవ్యరాశి సాంద్రతను అంచనా వేయండి. ఈ విలువలను అభ్యాసం 27లో పొందిన సోడియం పరమాణు సగటు ద్రవ్యరాశి సాంద్రతతో పోల్చండి.
సాధన:
కేంద్రకం సగటు ద్రవ్యరాశి = m
కేంద్రకంలో న్యూక్లియాన్ల సంఖ్య = n
కేంద్రకం యొక్క ద్రవ్యరాశి M = mA
కేంద్రకం వ్యాసార్థం r = roA1/3

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 36
m, ro లు స్థిరాంకాలు. అన్ని కేంద్రకాల సాంద్రత స్థిరం
m = 1.66 × 10-27 kg మరియు
ro = 1.2f = 1.2 × 10-15 m
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 37

అన్ని కేంద్రకాలకు p స్థిరం కనుక, ఇది సోడియం కేంద్రకం సాంద్రత కూడా అవుతుంది.
సోడియం కేంద్రకం సాంద్రత
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 38

ప్రశ్న 29.
లేజర్ (LASER) చాలా తీవ్రత కలిగిన, ఏకవర్ణక, ఏకదిశాత్మక కాంతి పుంజ జనకం. ఈ ధర్మాల ఆధారంగా సుదూరాలను కొలవడానికి లేజర్ను వినియోగించుకోవచ్చు. లేజర్ను ఒక కాంతి జనకంగా ఉపయోగించి భూమి నుంచి చంద్రుని దూరాన్ని చాలా ఖచ్చితంగా ఇదివరకే నిర్ణయించారు. చంద్రుడిపైకి కేంద్రీకరించిన ఒక లేజర్ కాంతి పుంజం చంద్రుని ఉపరితలం నుంచి పరావర్తనం చెంది భూమిని చేరడానికి 2.56s తీసుకొంటుంది. భూమి చుట్టూ పరిభ్రమించే చంద్రుడి కక్ష్యా వ్యాసార్ధం ఎంత?
సాధన:
ఇక్కడ t = 2.56sec
శూన్యంలో లేసర్ కాంతి వేగం = 3 × 108 m/s
లూనార్ కక్ష్యా వ్యాసార్థం, చంద్రుడు మరియు భూమికి మధ్యదూరం = x
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 39

ప్రశ్న 30.
సోనార్ (SONAR) (sound navigation and ranging) అనే సాధనం నీటి అడుగున ఉండే వస్తువులను వాటి స్థానాలను శోధించడానికి అతిధ్వని తరంగాలను ఉపయోగిస్తుంది. ఒక సబ్మెరైన్ (జలాంతర్గామి) లోని సోనార్ నుంచి వెలువడిన శోధక తరంగ ఉత్పత్తికి, శతృదేశ జలాంతర్గామిపై పరావర్తనం తరవాత పొందే ప్రతిధ్వనుల మధ్య కాల విలంబనం (time delay) 77.0 s అని కనుక్కొన్నారు. శతృదేశ సబ్మెరైన్ దూరం ఎంత ? (నీటిలో ధ్వని = 1450 ms-1).
సాధన:
ఇక్కడ t = 77.0 sec, υ = 1450 ms-1

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 40

ప్రశ్న 31.
ఇంతవరకూ ఆవిష్కరించిన, మన విశ్వంలోని అత్యంత దూరంలో గల వస్తువులు ఎంత దూరంలో ఉంటాయంటే అవి ఉద్గారించే కాంతి భూమిని చేరుకోవడానికి బిలియన్ల సంవత్సరాలు పడుతుంది. (క్వాజార్ (quasars) అని పిలిచే) ఈ వస్తువులకు అనేకమైన దిగ్రమ కలిగించే లక్షణాలుంటాయి. ఈ లక్షణాలను ఇంతవరకు ఎవరూ సంతృప్తికరమైన వివరణ ఇవ్వలేకపోయారు. ఒక క్వాజార్ నుంచి వెలువడే కాంతి మనల్ని చేరుకోవడానికి 3.0 బిలియన్ల సంవత్సరాలు పడితే km లలో దాని దూరం ఎంత?
సాధన:
పట్టుకాలం t = 3 బిలియన్ సంవత్సరాలు
= 3 × 109 సంవత్సరాలు
= 3 × 109 × 365 × 24 × 60 × 60 sec

శూన్యంలో కాంతివేగం, c = 3 × 108 m/s
= 3 × 105 m/s

దూరం = వేగం × కాలం
x = 3 × 105 × 3 × 109 × 365 × 24 × 60 × 60 km.
x = 2.84 × 1022 km

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 32.
సంపూర్ణ సూర్యగ్రహణం ఏర్పడినపుడు చంద్రుడు ఒక బిళ్లవలె సూర్యబింబాన్ని దాదాపుగా పూర్తిగా కప్పివేస్తాడని మనకు బాగా తెలిసిన వాస్తవం. దీని నుంచి, ఉదాహరణలు 3, 4 లోని సమాచారం ఆధారంగా చంద్రుడి ఉజ్జాయింపు వ్యాసాన్ని నిర్ణయించండి.
సాధన:
చంద్రుడి నుండి భూమి వరకు దూరం = 3.84 × 108 m
భూమి నుండి సూర్యుడి వరకు దూరం = 1.496 × 1011 m
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 41
సూర్యుడు AB వ్యాసం = 1.39 × 109 m
le ABE, ∆le CDEలు సర్వసమాన త్రిభుజాలు
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 42
CD = 3567.9 km ఇది చంద్రుడి వ్యాసం

ప్రశ్న 33.
ఈ శతాబ్దంలోని గొప్ప భౌతికశాస్త్రవేత్త అయిన P.A.M. Dirac ప్రకృతిలోని ప్రాథమిక స్థిరాంకాల సంఖ్యాత్మక విలువలతో ఆడుకోవడాన్ని ఎంతో ఇష్టపడేవాడు. ఈ ఆట అతణ్ణి ఒక ఆసక్తిదాయకమైన పరిశీలనవైపు నడిపించింది. పరమాణు భౌతికశాస్త్రం మూల స్థిరాంకాలు (c, e ఎలక్ట్రాన్ ద్రవ్యరాశి, ప్రోటాన్ ద్రవ్యరాశి), గురుత్వ స్థిరాంకం G నుంచి ‘కాలం’ మితిగల ఒక సంఖ్యను చేరుకోవచ్చని డిరాక్ కనుక్కొన్నాడు. పైగా, అది చాలా పెద్ద సంఖ్య. అంటే దాని పరిమాణం విశ్వం వయస్సుపై మనకున్న నేటి అంచనాకు (~ 15 బిలియన్ సంవత్సరాలు) చాలా దగ్గరగా ఉంటుందని తెలిసింది. ఈ పుస్తకంలో ఇచ్చిన ప్రాథమిక స్థిరాంకాలకు సబంధించిన పట్టిక నుంచి నీవు కూడా ఈ సంఖ్యను నిర్మించగలవేమో ప్రయత్నించి చూడు (లేదా మరేదైనా, నీవు ఆలోచించగలిగే ఆసక్తిదాయకమైన సంఖ్యను కనుక్కోవచ్చు).ఈ విశ్వం వయస్సుతో ఈ సంఖ్య ఏకీభవించడమనేది ఒకవేళ ఉపేక్షించదగనిది అయితే ప్రాథమిక స్థిరాంకాలకు ఉండే స్థిరత్వం (constancy) పట్ల ఇది ఏ అర్థాన్ని సూచిస్తుంది?
సాధన:
కాంతి వేగం, ఆ ఎలక్ట్రాన్ యొక్క ఆవేశం, mp ప్రోటాన్ ద్రవ్యరాశి, me ఎలక్ట్రాన్ ద్రవ్యరాశి, G విశ్వగురుత్వ స్థిరాంకం.
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 43
ఇది విశ్వం యొక్క వయస్సు.

సాధించిన సమస్యలు (Solved Problems)

ప్రశ్న 1.
కోణాలు (a) 1 (డిగ్రీ) (b) 1′ (చాపం యొక్క నిమిషం లేదా ఆర్మిన్), (c) 1″ (చాపం యొక్క సెకను లేదా ఆర్క్ సెకను) రేడియన్లలో లెక్కించండి. 360° = 2π rad, 1° = 60′ 1′ = 60″ లను ఉపయోగించండి.
సాధన:
(a) 360° = = 2π rad నుంచి
1° = (π/180) rad = 1.745 × 10-2

(b) 1° = 60′ = 1.745 × 10-2 rad
1’= 2.908 × 10-4 rad, 2.91 × 10-4 rad

(c) 1′ = 60″ = 2.908 × 104 rad
1″ = 4.847 × 10-6 rad, 4.85 × 10-6 rad.

ప్రశ్న 2.
ఒక వ్యక్తి తనకు సమీపంలో ఉండే శిఖరం (tower) దూరాన్ని అంచనా వేయాలను కొన్నాడు. శిఖరం C కు ఎదురుగా ఉండే బిందువు A వద్ద నిల్చొని, చాలా దూరంలో ఉండే వస్తువు 0 ను ACరేఖ వెంట ఉండటం గుర్తించాడు. అపుడు AC కి లంబ దిశలో 100 m దూరం బిందువు B వరకు నడిచాడు. తరువాత O, C ల వైపు మళ్ళీ చూశాడు. ౦ ఎక్కువ దూరంలో ఉంది కాబట్టి వాస్తవంగా BO దిశ AO దిశ ఒకటే అవుతాయి, కానీ, C దృష్టి రేఖ, మౌలికంగా ఉన్న దృష్టి రేఖ నుంచి θ = 40° కోణం విస్థాపనం చెందినట్లు గుర్తించాడు. (θను దృష్టి విక్షేపం అంటారు) అతని తొలిస్థానం A నుంచి టవర్ C దూరాన్ని అంచనా వేయండి.
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 44
సాధన:
పారలాక్స్ కోణం θ = 40°;
పటం నుంచి AB = AC tan θ
AC = AB/tan θ = 100 m/tan 40°
= 100 m/0.8391 = 119 m.

ప్రశ్న 3.
భూమి వ్యాసంపై ఉండే రెండు వ్యతిరేక బిందువులు A,B ల నుంచి, చంద్రుడిని పరిశీలించారు. చంద్రుడి వద్ద రెండు పరిశీలనా దిశలు ఏర్పరిచే కోణం 8 విలువ 1’54’. భూమి వ్యాసం సుమారుగా 1.276 × 107m అయితే, భూమి నుంచి చంద్రుని దూరాన్ని లెక్కించండి.
సాధన:
దత్తాంశం నుంచి θ = 1°54′ = 114′
= (114 × 60)” × (4.85 × 10-6) rad
1″ = 4.85 × 10-6 rad కాబట్టి,
θ = 3.32 × 10-2 rad,
అంతేగాక, b = AB = 1.276 × 107 m
భూమి -చంద్రుల మధ్య దూరం
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 45

ప్రశ్న 4.
సూర్యుడి కోణీయ వ్యాసం 1920″ అని కొలిచారు. భూమి నుంచి సూర్యుడి దూరం D విలువ 1.496 × 10″ m. అయితే సూర్యుడి వ్యాసం ఎంత?
సాధన:
సూర్యుడి కోణీయ వ్యాసం, α = 1920″
= 1920 × 4.85 × 10-6 rad
= 9.31 × 10-3 rad
సూర్యుడి వ్యాసం, d = α D
= (9.31 × 10-3) × (1.496 × 1011) m
= 1.39 × 109 m.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 5.
కేంద్రక పరిమాణాన్ని (10-15 నుంచి 10-14 m వ్యాప్తిలో) నిశితమైన పిన్ను కొనతో పోల్చితే, పరమాణు పరిమాణం సుమారుగా ఎంత ఉంటుంది ? పిన్ను కొన 10 m నుంచి 104 m వ్యాప్తిలో ఉన్నదనుకోండి.
సాధన:
కేంద్రక పరిమాణం 10-15 m నుంచి 10-14 m వ్యాప్తిలో ఉంది. నిశితమైన పిన్ను చివర 10-5 m నుంచి 10-4 m వ్యాప్తిలో ఉన్నదని అనుకొంటే మనం పరిమాణ విలువను 1010 కారకంతో పెంచుతున్నామన్నమాట. కాబట్టి పరమాణువుకు 10-10 m (దాదాపు) పరిమాణం ఉన్నప్పుడు దానిని 1 m పరిమాణానికి పెంచినామన్న మాట. అంటే పరమాణువు లోపలి కేంద్రకం పరిమాణం దాదాపు 1 మీటరు పొడవు వ్యాసార్ధం ఉన్న గోళం కేంద్రం వద్ద ఉంచిన నిశిత పిన్ను కొన అంత చిన్నదిగా ఉంటుంది.

ప్రశ్న 6.
ఒక జాతీయ పరిశోధనాశాలలోని ప్రామాణిక గడియారంతో పోలుస్తూ రెండు గడియారాల పనితీరును పరీక్షిస్తున్నారు. ప్రామాణిక గడియారం మధ్యాహ్నం 12 : 00:00 సమయాన్ని సూచిస్తున్నప్పుడు ఆ రెండు గడియారాల్లోని రీడింగులు ఈవిధంగా ఉన్నాయి.

గడియారం 1గడియారం 2
సోమవారం12:00:0510:15:06
మంగళవారం12:01:1510:14:59
బుధవారం11:59:0810:15:18
గురువారం12:01:5010:15:07
శుక్రవారం11:59:1510:14:53
శనివారం12:01:3010:15:24
ఆదివారం12:01:1910:15:11

ఖచ్చితత్వం కలిగిన కాలవ్యవధుల కొలతలు అవసరమయ్యే ప్రయోగాన్ని నీవు చేస్తున్నావని అనుకొంటే, పై రెండు గడియారాల్లో దేన్ని ఎంచుకొంటావు?
సాధన:
ఏడు రోజులపాటు చేసే పరిశీలనల్లో మొదటి గడియారం (1) సమయాల్లో వచ్చే మార్పుల వ్యాప్తి 162 5 కాగా, రెండో గడియారానికి (2) 31 5 అవుతుంది. గడియారం (1) చూపే సగటు రీడింగ్, గడియారం (2) చూపే సగటు రీడింగ్ కంటే ప్రామాణిక కాలానికి చాలా దగ్గరగా ఉంది. ఇక్కడ గమనించవలసిన ముఖ్యమైన అంశం ఏమంటే ఖచ్చితత్వం అవసరమయ్యే ప్రయోగాలకు గడియారం చూపే వ్యాప్తియే ప్రధానం కాని అది చూపే శూన్యదోషం కాదు. ఎందుకంటే, ‘శూన్యదోషాన్ని’ ఎప్పుడైనా సులువుగా సవరించుకోవచ్చు. కాబట్టి గడియారం (1) కంటే రెండో గడియారాన్నే ఎంచుకోవాల్సి ఉంటుంది.

ప్రశ్న 7.
లఘులోలకం డోలనావర్తన కాలాన్ని కొలుద్దాం. వరుస కొలతల్లో వచ్చిన రీడింగ్లు 2.63s, 2.56, 2.42 s, 2.71, చివరగా 2.80s. పరమదోషాలను, సాపేక్షదోషం లేదా దోష శాతాన్ని లెక్కించండి.
సాధన:
లోలకం సగటు డోలనావర్తన కాలం
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 46

ప్రశ్న 8.
థర్మామీటరుతో రెండు వస్తువుల ఉష్ణోగ్రతలను t1 = 20 °C ± 0.5 °C, t2 = 50°C ± 0.5°C గా కొలిచారు. వాటి ఉష్ణోగ్రతా భేదాన్ని, దానిలోని దోషాన్ని లెక్కించండి.
సాధన:
t’ = t2 – t1
= (50 °C ± 0.5 °C) – (20 °C ± 0.5 °C)
t’ = 30 °C ± 1 °C.

ప్రశ్న 9.
నిరోధం R = V/I ఇందులో V = (100 ± 5) V, I = (10 ± 0.2) A. అయితే లోని దోషశాతాన్ని కనుక్కోండి.
సాధన:
V లోని దోష శాతం 5, అలాగే I లో దోష శాతం 2.
కాబట్టి R లో మొత్తం దోషం 5 + 2 = 7%.

ప్రశ్న 10.
రెండు నిరోధకాల నిరోధాలు R1 = 100 ± 3 ohm, R2 = 200 ± 4 ohm. వీటిని (a) శ్రేణిలో, (b) సమాంతరంగా కలిపారు. (a) శ్రేణీ సంయోగానికి, (b) సమాంతర సంయోగానికి తుల్య నిరోధాలను కనుక్కోండి. సూచన :
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 47
సాధన:
a) శ్రేణి సంయోగానికి తుల్య నిరోధం
R = R1 + R2 = (100 ± 3)
ohm + (200 ± 4) ohm = 300 ± 7 ohm

b) సమాంతర సంయోగానికి తుల్య సంబంధం
AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత 48
అప్పుడు R’ = 66.7 ± 1.8 ohm
(ఇక్కడ ∆R విలువను, 2గా సూచించడానికి బదులు 1.8 గానే వ్యక్తం చేశాం. `ఎందుకంటే, సార్థక సంఖ్యల నియమాలకు అనుగుణంగా ఉండటం కోసం).

ప్రశ్న 11.
Z = A4B1/3/CD3/2 అయితే 2 లో సాపేక్ష దోషాన్ని కనుక్కోండి.
సాధన:
Z లో సాపేక్ష దోషం, ∆Z/Z = 4(∆A/A) + (1/3) (∆B/B) + (∆C/C) + (3/2) (∆D/D).

ప్రశ్న 12.
లములోలకం డోలనావర్తన కాలం = 2π √L/g . 1 mm తెలిసిన యదార్థతతో కొలచిన L విలువ 20.0 cm. 100 డోలనాలకు పట్టిన కాలాన్ని 1s పృథక్కరణం ఉన్న చేతి గడియారంతో 90 s అని కనుక్కొన్నారు. అయితే g విలువను నిర్ణయించడంలో యదార్ధత ఎంత?
సాధన:
g = 4π²L/T²; ఇక్కడ T = \(\frac{t}{n}\), ∆T = \(\frac{\Delta t}{n}\)
కాబట్టి, \(\frac{\Delta T}{T}=\frac{\Delta t}{t}\) ఈ L, t రెండింటిలోని దోషాలు
కనీసపు కొలత దోషాలు కాబట్టి
(∆g/g) = (∆L/L) + 2(∆T/T)
= \(\frac{0.1}{20.0}\) + 2(\(\frac{1}{90}\)) = 0.027
అందువల్ల g లోని దోషశాతం
100 (∆g/g) = 100(∆L/L) + 2 × 100 (∆T/T) = 3.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 13.
ఒక ఘనం యొక్క ఒక్కొక్క భుజం పొడవును 7.203 m గా కొలిచారు. దాని మొత్తం ఉపరితల వైశాల్యం, ఘనపరిమాణాల విలువలను తగిన సార్ధక సంఖ్యల వరకు కనుక్కోండి.
సాధన:
కొలిచిన పొడవులో నాలుగు సార్థక సంఖ్యలు ఉండటం వల్ల మనం లెక్కించే వైశాల్యం, ఘనపరిమాణాలను కూడా నాలుగు సార్ధక సంఖ్యల వరకే సవరించవలసి ఉంటుంది.
ఘనం ఉపరితల వైశాల్యం = 6(7.203)² m²
= 311.299254 m²
= 311.3 m²

ఘనం ఘనపరిమాణం = (7.203)³ m³
= 373.714754 m³
= 373.7 m³.

ప్రశ్న 14.
5.74 g పదార్థం 1.2 cm3 ఘనపరిమాణం ఆక్రమిస్తుంది. సార్ధక సంఖ్యలను దృష్టిలో ఉంచుకొని దాని సాంద్రత విలువను వ్యక్తపరచండి.
సాధన:
కొలచిన ద్రవ్యరాశిలో మూడు సార్ధక సంఖ్యలు ఉంటే కొలచిన ఘనపరిమాణంలో రెండే సార్థక సంఖ్యలు ఉన్నాయి. కాబట్టి, సాంద్రత విలువను 2 సార్థక సంఖ్యల వరకు మాత్రమే వ్యక్తపరచాలి.
సాంద్రత = \(\frac{5.74}{1.2}\) g cm-3 = 4.8 g cm-3.

ప్రశ్న 15.
\(\frac{1}{2}\)mυ² = mgh అనే ఒక సమీకరణాన్ని పరిగణిద్దాం. ఇందులో m అనేది వస్తువు ద్రవ్యరాశి, υ దాని వేగం, g గురుత్వ తరణం, h దాని ఎత్తు. ఈ సమీకరణం మితీయంగా సరియైందో కాదో పరీక్షించి చూడండి.
సాధన:
ఎడమవైపు (LHS) ఉన్న మితులు
[M] [LT-1]² = [M] [L²T-2] = [M L²T-2]
కుడివైపు (RHS) ఉన్న మితులు
[M] [LT-2] [L] = [M] [L²T-2] = [M L²T-2]
LHS, RHS లలోని మితులు ఒకటే కాబట్టి సమీకరణం మితీయంగా సరైనది.

ప్రశ్న 16.
శక్తికి SI ప్రమాణం J = kg m²s-2; అదేవిధంగా వడి υ, త్వరణం a కు ప్రమాణాలు వరుసగా ms-1, ms-2. అయితే గతిజశక్తి (K) కి కింద ఇచ్చిన ఫార్ములాల్లో మితీయంగా దేనిని క్రమ విరుద్ధమైనదిగా తోసిపుచ్చుతావు? (m వస్తువు ద్రవ్యరాశిని సూచిస్తుంది)
(a) K = m²υ³ (b) K = (1/2) mυ²
(c) K = = ma (d) K = (3/16) mυ²
(e) K = (1/2) mυ² + ma.
సాధన:
ప్రతి సరైన సమీకరణం లేదా ఫార్ములాలో ‘సమీకరణానికి రెండువైపులా ఒకే మితులు ఉండాలి. అదేవిధంగా ఒకే రకమైన మితులుండే రాశులను మాత్రమే సంకలనం లేదా వ్యవకలనం చేయాలి. కుడివైపు గల రాశికి మితులు (a) విషయంలో[M²L³T-3], (b), (d) లకు [ML²T-2] (c)కి [MLT-2], (e) కి కుడివైపునున్న రాశికి సరియైన మితులు లేవు. ఎందుకంటే భిన్న మితులు కలిగిన రెండు రాశులను కలపడమైంది. గతిజ శక్తి Kకి ఉండే మితులు [ML²T-2] కాబట్టి (a), (c), (e) ఫార్ములాలను తోసిపుచ్చవలసిందే. అయితే, మిగతా రెండు ఫార్ములాలు (b) లేదా (d) లలో ఏది సరియైనదో మితుల ద్వారా చెప్పలేం. అందుకని, గతిజ శక్తికి ఇచ్చిన (అధ్యాయం లో) నిజ నిర్వచనం ప్రకారం, గతిజ శక్తికి సరియైన ఫార్ములాను (b) సూచిస్తుంది.

AP Inter 1st Year Physics Study Material Chapter 2 ప్రమాణాలు, కొలత

ప్రశ్న 17.
దారానికి గుండు తగిలించిన లఘులోలకాన్ని పరిగణించండి. గురుత్వ బల చర్య వల్ల ఇది డోలనాలను చేస్తుంది. లోలకం డోలనావర్తన కాలం(T), పొడవు(1), గుండు ద్రవ్యరాశి (m), గురుత్వ త్వరణం(g) లపై ఆధారపడుతుందను కోండి. ‘మితుల పద్ధతిని ఉపయోగించి దాని ..డోలనావర్తన కాలానికి సమాసాన్ని ఉత్పాదించండి.
సాధన:
l, g, m రాశులపై డోలనావర్తన కాలం T ఆధారపడటాన్ని వాటి లబ్ధంగా ఇలా రాయవచ్చు.
T = k lx gy mz

ఇక్కడ k మితిరహిత స్థిరాంకం, x, y, z లు ఘాతాంకాలు. ఇరువైపులా మితులను తీసుకొంటే,
[L°M°T¹] = [L¹]x [L¹T-2]y [M¹]z = Lx + y T-2y Mz
ఇరువైపులా ఉన్న మితులను సమానం చేస్తే,
x + y = 0; -2y = 1; z = 0
అందువల్ల, x = \(\frac{1}{2}\), y = –\(\frac{1}{2}\), z = 0
అప్పుడు T = kl1/2 g-1/2 లేదా T = k\(\sqrt{\frac{l}{g}}\)

k విలువను మితుల పద్ధతి ద్వారా కనుక్కోలేమని గమనించండి. ఇక్కడ కుడివైపు ఉన్న ఫార్ములాను ఏదో ఒక సంఖ్యతో గుణించినంత మాత్రాన ఎలాంటి ప్రభావం ఉండదు. ఎందుకంటే ఆ సంఖ్య దాని మితులను ప్రభావితం చేయదు.
నిజానికి k = 2π. అందువల్ల,
T = 2π \(\sqrt{\frac{l}{g}}\)

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature

Long Answer Questions

Question 1.
Explain the composition, powers, and functions of the State Legislative Assembly.
Answer:
The Constitution provides for a Legislature for every State on the model of the Parliament. As per Article 168, the State Legislature consists of the Governor and one or two Houses. In India, while some States have Bicameral Legislatures, others have Unicameral Legislatures. Andhra Pradesh at present possesses Unicameral Legislature.

The Lower House of the State Legislature is known as the Legislative Assembly or Vidhana Sabha and the Upper House is the Legislative Council or Vidhana Parishad.

Composition of Legislative Council (Vidhana Parishad) :
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than \(\frac{1}{3}\)rd of the total membership of the State Assembly. The Legislative Council consists of both nominated and elected members.

The election is conducted through the indirect method by means of proportional representation with a single transferable vote.

Distribution of Seats :

  1. \(\frac{1}{3}\)rd are elected by the members of the State Assembly.
  2. \(\frac{1}{3}\)rd are elected by members of local bodies.
  3. \(\frac{1}{12}\)th are elected by teachers.
  4. \(\frac{1}{12}\)th are elected by graduates.
  5. The remaining \(\frac{1}{6}\)th members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Qualifications :
The members of the Council 1) must be citizens of India, 2) must have completed 30 years of age and 3) must possess such other qualifications as may be prescribed by the Legislature.

Term :
the members are elected for a period of 6 years. But \(\frac{1}{3}\)rd of them retire for every 2 years. The Council is a permanent body. It cannot be dissolved by the Governor.

Chairman and Deputy Chairman:
The Council has Chairman and a Deputy Chairman who are elected by the members of the Council from among themselves. The Chairman presides over the meetings of the Council.

Legislative Assembly (Vidhana Sabha) :
Composition of the Legislative Assembly :
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. It’s maximum strength is fixed at 500 and minimum strength at 60. Only the Legislative Assembly of Sikkim has less than 60 because of it’s small population. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of Office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding Officers :
The Presiding officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the Speaker.

Powers and Functions of the State Legislature :
The State Legislature has the following powers.

1) Legislative Powers and Functions :
The State Legislature has the power to make laws on all the subjects included in the State List. It has also the power to make laws in respect of subjects included in the Concurrent List. However, such a law should not disagree with a law already made by the Parliament on the same subject. In the making of laws, Legislative Assembly has been given more powers than the Legislative Council. The Legislative Council at the most may delay the legislation for a period of 4 months. Later the Assembly sends the bill to the Governor for his assent.

2) Constitutional Powers and Functions :
Even though, the Legislature has no powers to move the Constitution amendment bills, it’s consent is required for amending certain provisions of the Constitution. Such bills have to be referred to it after they are approved by the Parliament.

3) Executive Powers and Functions :
The State Legislature exercises control over the Council of Ministers. It’s members make the Ministers individually and collectively responsible to the Legislature. The Council of Ministers is collectively responsible to the State Legislative Assembly. The Legislature can expose the actions of Executive, through questions, debates and adjournment motions. In controlling the Executive, the Legislative Assembly has more powers than the Council. The Ministry has to resign when the Legislative Assembly passes no confidence motion against the Government.

4) Financial Powers and Functions:
The State Legislature exercises complete control over the finances of the State Government. It sanctions money to the State Government to enable it to run the administration. It may pass, reduce or reject the demands for grants presented to it by the Government. It may accept or reject proposals for taxation and borrowings presented to it by the Government. In financial matters the Assembly is more powerful than the Council. Because all money bills, including the Budget, shall be introduced first only in the Assembly. It can accept or reject any recommendations made by the Council.

5) Electoral Powers :
The elected members of the Assembly participate in the election of the President. They also elect the representatives of the State to the Rajya Sabha and l/3rd members of the Legislative Council if the State Legislature is bicameral. They also elect the presiding officers and deputy presiding officers of Assembly and Council.

Miscellaneous Powers : The state legislature :

  1. Safeguards the dignity and privileges of its members.
  2. Suspends, expels or terminates their membership.
  3. Examines the report of the State Public Service Commission and the Comptroller and Auditor General etc.

Conclusion:
The State Legislature plays an important role in the State Administration. It makes necessary laws for the welfare of the people of the State. It controls the Executive by making it responsible for their actions.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write briefly the composition, powers and functions of the. State Legislative Council.
Answer:
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than \(\frac{1}{3}\)rd of the total membership of the State Assembly. The Legislative Council is a body of partly nominated and partly elected members.

The election is conducted through indirect method by means of proportional representation with a single transferable vote.
Composition of the Council:

  1. \(\frac{1}{3}\)rd members are elected by the Legislative Assembly.
  2. \(\frac{1}{3}\)rd members are elected by members of local bodies.
  3. \(\frac{1}{12}\)th members are elected by teachers.
  4. \(\frac{1}{12}\)th members are elected by graduates.
  5. The remaining \(\frac{1}{6}\)rd members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Tenure :
The Legislative council is a quasipermanent House. l/3rd of the members of this House retire every two years. But the term of each member is six years. New members are elected in the place of retired members. All the members of the House do not retire at a time as it is a permanent House.

Powers and Functions of State Legislative council:
State Legislative council is the primary law making body along with the Legislative Assembly. The State Legislative Council has the following powers and Functions.

a) Legislative powers and Functions :
The Legislative Council does not possess equal powers and functions when compared to counterpart, State Legislative Assembly. It is said that the Legislative Council enjoys equal status and not power. However it exercises the following powers and functions. All the bills, other than money bills may be introduced in either of the House. They will be sent to the assent of the Governor only with the approval of both the Houses. The Council may reject any bill and spnt it back for the reconsideration of the Assembly. However, incase of a disagreement between two Houses, the decision of the Assembly will be supreme.

The Council must approve all the bills sent by the Assembly with in a period of three months or at the maximum of four months. It implies that the Council can withhold its assent over the bills sent by the Assembly for a maximum period of four months. Thus, the Legislative Council can only delay the bills but the Legislative Assembly can override it.

b) Executive Powers and Functions :
The State Legislative Council has very limited executive powers when compared to that of the Assembly. The council of Ministers headed by the Chief Minister is responsible for its acts only to the Assembly and not to the Council. The Council cannot decide the future of the Council of Ministers. However, the Council can influence the policies and programmes of the ministers by asking questions and supplementary questions by drawing the call attention motion etc., but they cannot force the Council of Ministers to resign.

c) Financial Powers and Functions :
The Legislative Council has only limited powers in the financial matters. Money bills cannot at first be introduced in the Legislative Council first. The Council must accept all money bills with or without recommendation within fourteen days of the receipt of the bill. The Assembly possess the discretion powers either to accept or reject these recommendations. If the Council does not return the Money Bill to the Assembly within 14 days, then the bill is deemed to have been passed by both the Houses. It is clear that in the financial field the Legislative Council has a subordinate status and that Legislative Assembly has dominant position.

d) Electoral Functions:
The Legislative Council elects a Chairman and Deputy Chairman to preside over its meetings in a dignified manner. Some of its members are elected to various legislative committees like Public Accounts Committee, Estimates Committee and Public Undertakings committee etc.

e) Other Functions :
The Legislative council acts as the best means for formulating and consolidating public opinion. It discusses technical and other contemporary matters, as there are experts in various fields.

Question 3.
Explain the role and responsibilities of the Speaker of Legislative Assembly.
Answer:
The members of State Legislative Assembly elect one among them as a speaker to conduct the business of the House. His term of office is five years.

The speaker is the guardian of the Rights and Liberties of the members of the House.

Role and Responsibilities (or) Powers and Functions of the Speaker :
The powers and functions of the speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of casting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent and incriminatory references from the records.
  9. He allows the members to speak in the House.
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committee of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Short Answer Questions

Question 1.
Write a note on the Legislative Assembly.
Answer:
Legislative Assembly is the Lower House of the State Legislature. The Members of Legislative Assembly are called M.L.As. According to Article 170of the Indian Constitution it consists of not more than 500 members and not less than 60 members. It means that it’s strength depends on the population and size of the state. But small states have been allowed to have less number of members. Thus Goa and Mizoram have only 40 members, while Sikkim has 32 Members.

Composition of the Legislative Assembly:
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding officers :
The Presiding Officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the speaker.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write a note on Estimates Committee.
Answer:
According to the Rules of Procedure and Conduct of Business in the State Legislature, the Estimates committee consists of 20 members. Among them 15 members belong to Assembly. The remaining 5 members belong to Legislative Council. The members hold office for a period of one year. They are elected through in indirect election.

Functions :
The functions of the Estimates Committee in the State Legislature are the same as that of Estimates Committee of Lok Sabha. These are given here under.

  1. Estimates Committee exercises control over public expenditure.
  2. It suggests fiscal reforms in organization, the efficiency or administration reforms consistent with the policy underlying estimates.
  3. It advises alternative policies for securing efficiency and economy in administration.
  4. It examines whether the money is well laid out within the limits of the policy implied in the estimates.
  5. It also suggests the form in which the estimates shall be presented to the Assembly.

Question 3.
What do you know about Public Accounts Committee? [Mar.-18, 17]
Answer:
Public Accounts Committee consists of 20 members out of which 15 members belong to Assembly and 5 members belong to Legislative Council. They are elected through indirect election by following the principle of proportional representation Jor a period of one year. The Chairman is normally the member of Opposition Party. The Ministers of Cabinet cannot be member of Public Accounts Committee.

Functions :
Public Account Committee performs the following functions.

  1. The committee examines the accounts showing the appropriation of sums granted by the house for expenditure of the state government.
  2. It scrutinizes the appropriation accounts of the state and the reports of the Comptroller and Auditor General.
  3. It shall be the duty of the Public Accounts Committee to examine such a trading, manufacturing, and profit and loss accounts and balance sheets and the accounts of the state government and also to consider the report of the Comptroller and Auditor General.
  4. The committee carefully considers the accounting and audit procedures.
  5. The committee is not concerned with the question of policy approved by the legislature.
  6. The committee investigates expenditure after it has already incurred. An overall, this committee is generally described as a ‘post-mortem committee’.

Question 4.
Write the powers and functions of Vidhcma Sabha Speaker. [Mar. 16]
Answer:
The powers and functions of the Speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of easting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent arid incriminatory references from the records.
  9. He allows the members to speak in the House. ‘
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committees of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Very Short Answer Questions

Question 1.
Qualifications of M.L.A. [Mar. 18, 16]
Answer:
A person who wishes to contest for the membership of the State Legislative Assembly must be possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed the age of 25 years.
  3. He should possess such other qualifications as prescribed by any act of Parliament.
  4. However, no person can simultaneously be a member of any House of the Parliament and of a State Legislature.

Question 2.
Qualifications of M.L.C.
Answer:
A person who wishes to contest for the membership of the State Legislative Council must possess the following qualifications.
a) He should be a citizen of India.
b) He should have completed 30 years of age.
c) He should possess such other qualifications as laid down by an Act of Parliament.

Question 3.
Quorum.
Answer:
Quorum is the minimum number of members required to be present in the house before it can transact any business. According to Article 188 of the constitution, the Quorum for conducting,the State Legislative Assembly meeting was fixed at 1/10th of the total membership. However, in some states, where the strength of the State Legislative Assembly is very less, the quorum will be a minimum number of 10. The speaker decides whether there is a quorum or not on a particular day.

Question 4.
Salaries and Allowances of M.L.A.
Answer:
The salary of MLA is decided by the respective State Legislature as per the Article 164 of the Indian constitution. The members of Andhra Pradesh State Legislative Assembly receive a monthly salary of ₹ 90,000/- which includes a basic pay of ₹ 15,000/- and constituency allowance of ₹ 75,000/-. Those legislators who are not provided government accommodation will get an additional ₹ 10,000/- as H.R.A members also get daily allowance of ₹ 800/- when the state legislature is in session.

Question 5.
Privileges of State Legislature.
Answer:
Privileges of a State Legislature are a sum of special rights, immunities and exemptions enjoyed by the State Legislatures. They are necessary in order to secure independence and effectiveness of their actions. The Houses cannot maintain the authority, dignity and honour without these privileges. They can protect their members from any obstructions in the discharge of their legislative responsibilities.

  1. Collective privileges
  2. Individual privileges

i) Collective Privileges :
The legislature has the right to publish its reports, debates and proceedings and also to prohibit others publishing the same.

ii) Individual Privileges:
The privileges belonging to the members of state legislature individually. They can not be arrested during the session of the state legislature or 40 days before and after the end of the session.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 6.
Brief History of AP legislature.
Answer:
The Andhra state was formed on October 1, 1953 Andhra State legislature initially had 140 MLAs. Elections were held to the Andhra State Legislative Assembly for the first time in 1955.

As per the recommendations of states re-organization committee, Hyderabad State was merged with Andhra State on linguistic basic and formed into Andhra Predesh State which had 245 MLAs [Including 150 MLAs of Hyderabad State.] Elections were held to the Andhra Pradesh legislative Assembly in 1957.

The State Legislative Council was established on July 1,1958. Since then it continued to exist till June 1, 1985, before being abolished. Again on March 30, 2007 the Andhra Pradesh Legislature became again bicameral after the revival of the legislative council.

Question 7.
Chairman of Legislative council. [Mar. 16]
Answer:
There will be a chairman in the Legislative council for conducting the meetings. He is elected by the members of the Legislative council among themselves. Dr. A. Chakrapani Yadav is the Present Chairman of Legislative council of Andhra Pradesh.

Question 8.
Deputy Speaker.
Answer:
The members of State Legislative Assembly elect one among them as a Deputy Speaker. His term of office is 5 years. The deputy speaker performs the duties in the absence of the speaker.

Question 9.
Deputy Chairman of Legislative Council.
Answer:
The Members of State Legislative Council elect one among them as a Deputy Chairman. The Deputy Chairman performs the duties in the absence of the Chairman.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 10.
Types of committees. [Mar. 18]
Answer:
The committees are of two types i.e., Standing committees and Ad-hoc committees.
i) Standing Committees:
Standing committees deal with specific business (financial matters) Ex : Estimates committee, Public accounts committee and Committee on public undertakings.

ii) Ad-hoc Committees :
Ad-hoc committees are concerned with the matters of temporary nature. They cease to exist after completion of the work. They perform some specific functions assigned to them from time to time.

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c)

I.

Question 1.
Express x dy – ydx = \(\sqrt{x^2+y^2}\) dx in the form F (\(\frac{y}{x}\)) = (\(\frac{dy}{dx}\)).
Solution:
x. dy – y dx = \(\sqrt{x^2+y^2}\) dx
\(\frac{dy}{dx}\) – y = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 1

Question 2.
Express (x – y Tan-1 \(\frac{y}{x}\))dx + x Tan-1\(\frac{y}{x}\) dy = 0 in the form F(\(\frac{y}{x}\)) = \(\frac{dy}{dx}\).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 2

Question 3.
Express x\(\frac{dy}{dx}\) = y(log y – log x + 1) in the from F (\(\frac{y}{x}\)) = \(\frac{dy}{dx}\)
Solution:
x . \(\frac{dy}{dx}\) = y(log y – log x + 1)
\(\frac{dy}{dx}\) = \(\frac{y}{x}\)(log\(\frac{y}{x}\) + 1)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Solution:
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Put y = vx
\(\frac{dy}{dx}\) = v + x\(\frac{dv}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 3

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 2.
(x² + y²) dy = 2xy dx
Solution:
(x² + y²) dy = 2xy dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 4
1 + v² = A(1 – v²) + Bv(1 – v) + Cv(1 + v)
v = 0 ⇒ = A
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
v = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = -2B
B = -1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 5

Question 3.
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Solution:
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 6
Multiplying with (v + 1)³
3 + v² = A(v + 1)² + B(v + 1) + C
v = – 1 ⇒ 3 + 1 = C ⇒ C = 4
Equating the co-efficients of v²
A = 1
Equating the co-efficients of v
0 = 2A + B
B = -2A = -2
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 7

Question 4.
y²dx + (x² – xy)dy = 0
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 8
v – log v = log x + log k
v = log v + log x + log k
= log k (vx)
\(\frac{y}{x}\) = log ky
Solution is ky = ey/x

Question 5.
\(\frac{dy}{dx}=\frac{(x+y)^2}{2x^2}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 9

Question 6.
(x² – y²)dx – xy dy = 0
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 10
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 11

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 7.
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Solution:
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 12
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 13

Question 8.
y²dx + (x² – xy + y²) dy = 0
Solution:
y²dx = – (x² – xy + y²)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 14
1 – v + v² = A(1 + v²) + (Bv + C)v
v  =  0 ⇒ 1 = A
Equating the co-efficients of v²
1 = A + .B ⇒ B = 0
Equating the co-efficients of v
-1 = C
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 15

Question 9.
(y² – 2xy)dx + (2xy – x²)dy = 0
Solution:
(y² – 2xy)dx + (2xy – x²)dy = 0
(2xy – x²)dy = – (y² – 2xy)dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 16
2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 17

Question 10.
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 18
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 19

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\)
Solution:
x dy – y dx = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 20

Question 12.
(2x – y)dy = (2y – x)dx
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 21
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 22
⇒ (y – x)² = c²(y + x)²(y² – x²)
⇒ y – x = c²(y + x)³
⇒ (x + y)³ = c(x – y) where c = \(\frac{-1}{c^2}\)
∴ (x + y)³ = c(x – y)

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 13.
(x² – y²)\(\frac{dy}{dx}\) = xy
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 23
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 24

Question 14.
2\(\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{x^2}\)
Solution:
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 25
(y – x)² = y²c²x
Solution is y²x = c(x – y)²

III.

Question 1.
Solve : (1 + ex/y)dx + ex/y(1 – \(\frac{x}{y}\))dy = 0.
Solution:
Put x = vy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 26

Question 2.
Solve : x sin \(\frac{y}{x}.\frac{dy}{dx}\) = y sin \(\frac{y}{x}\) – x
Solution:
Dividing with x, we have
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 27

Question 3.
Solve : x dy = (y + x cos² \(\frac{y}{x}\))dx
Solution:
x.\(\frac{dy}{dx}\) = y + xcos² \(\frac{y}{x}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 28

Question 4.
Solve : (x – y log y + y log x)dx + x(log y – log x)dy = 0
Solution:
Dividing with x. dx we get
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 29
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 30

Question 5.
Solve : (y dx + x dy) x cos \(\frac{y}{x}\) = (x dy – y dx) y sin \(\frac{y}{x}\).
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 31
∴ This is a homogeneous equation
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 32
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 33

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 6.
Find the equation of a curve whose gradient is \(\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 34
tan v = – log |x| + c
This curve passes through (1, \(\frac{\pi}{4}\))
tan \(\frac{\pi}{4}\) = c – log 1
c = 1
∴ Equation of the curve is
tan v = 1 – log |x|
tan \(\frac{y}{x}\) = 1 – log |x|

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(c)

I. Evaluate the following integrals.

Question 1.
∫x sec² x dx on I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n is integer}.
Solution:
∫uvdx = u∫vdx – ∫[\(\frac{d}{du}\)(U).∫vdx]dx
Let v = sec² x and u = x then
∫x sec² x dx = x(tan x) – ∫tan x dx
= x tan x – log|sec x| + C

Question 2.
∫ex(tan-1 x + \(\frac{1}{1+x^2}\))dx, x ∈ R.
Solution:
∫ex[f(x) + f'(x)] dx = ex. f(x) + C
Let f(x) = tan-1 x so that f'(x) = \(\frac{1}{1+x^2}\)
∴ ∫ex(tan-1 x + \(\frac{1}{1+x^2}\))dx = extan-1 x + C

Question 3.
∫\(\frac{log x}{x^2}\) dx on (0, ∞).
Solution:
∫\(\frac{log x}{x^2}\) dx = (log x)(-\(\frac{1}{x}\)) + ∫\(\frac{1}{x}\) . \(\frac{1}{x}\)dx
= – \(\frac{1}{x}\)log x – \(\frac{1}{x}\) + C

Question 4.
∫(log x)² dx (0, ∞).
Solution:
∫(log x)² dx = (log x)² x – ∫x . 2log x.\(\frac{1}{x}\) dx
= x (log x)² – 2 ∫log x dx
= x (log x)² – 2(x . log x – ∫x\(\frac{1}{x}\)dx)
= x(log x)² – 2x. log x + 2x + c

Question 5.
∫ex(sec x + sec x . tan x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
∫ex(sec x + sec x . tan x)dx = ex.sec x + C
[∫ex[f(x) + f'(x)]dx = ex f(x) + C]

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫ex cos x dx on R.
Solution:
I = ∫ex cos x dx = ex sin x – ∫sin x . ex dx
= ex. sin x + ex. cos x – ∫ex . cos x dx
= ex(sin x + cos x) – I
2I = ex(sin x + cos x)
I = \(\frac{e^x}{2}\)(sin x + cos x) + C

Question 7.
∫ex(sin x + cos x) dx on R.
Solution:
∫ex(sin x + cos x) dx
f(x) = sin x ⇒ f'(x) = cos x
∴ ∫ex(sin x + cos x) dx = ex. sin x + C

Question 8.
∫(tan x + log sec x)ex dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π) n ∈ Z.
Solution:
t = log|sec| ⇒ dt = \(\frac{1}{\sec x}\). sec x .tan x dx
= tan x dx
∫(tan x + log sec x)ex dx = ex.log|sec x| + C

II. Evaluate the following integrals.

Question 1.
∫xnlog x dx on (0, ∞), n is a real nember and n ≠ -1.
Solution:
∫xnlog x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 1

Question 2.
∫log (1 + x²) dx on R.
Solution:
∫log (1 + x²) dx on
= [log (1 + x²). x – ∫x\(\frac{1}{1+x^2}\)2x dx
= x log (1 + x²) – 2∫\(\frac{1+x^2-1}{1+x^2}\)dx
= x log (1 + x²) – 2∫dx + 2∫\(\frac{dx}{1+x^2}\)
= x log (1 + x²) – 2x + 2 tan-1 x + C

Question 3.
∫\(\sqrt{x}\) log x dx on (0, ∞).
Solution:
∫\(\sqrt{x}\) log x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 2

Question 4.
∫e√x dx on (0, ∞).
Solution:
t = √x ⇒ x = t²
dx = 2t dt
∫e√x dx = 2∫t et dt
= 2 [t et – ∫et dt
= 2 (t et – et) + C
= 2√x e√x – 2e√x + C

Question 5.
∫x² cos x dx on R.
Solution:
∫x² cosx dx = x²(sin x) – ∫sin x(2x dx)
= x² sin x + 2∫x(-sin x)dx
= x². sin x + 2[x cos x – ∫cos x dx]
= x² sin x + 2x cos x – 2 sin + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫x sin² x dx on R.
Solution:
∫x sin² x dx = \(\frac{1}{2}\)∫x(1 – cos 2x) dx
= \(\frac{1}{2}\)[∫x dx – ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) – {x . \(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x + \(\frac{1}{4}\)∫sin 2x dx
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x – \(\frac{1}{8}\)cos 2x + C

Question 7.
∫x cos² x dx on R.
Solution:
∫x cos² x dx = \(\frac{1}{2}\)∫x(1+ cos 2x)dx
= \(\frac{1}{2}\)[∫x dx + ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) + {x.\(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x – \(\frac{1}{4}\) ∫sin 2x dx
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x + \(\frac{1}{8}\)cos 2x + C

Question 8.
∫cos √x dx on R.
Solution:
t = x = t² ⇒ dx = 2t dt
I = 2∫t. cos t dt = 2 (t sin t – ∫sin t dt)
= 2(t sin t + cos t) + C
= 2√x sin √x + 2 cos√x + C

Question 9.
∫x sec² 2x dx on I ⊂ R\ {(2nπ + 1)\(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫x sec² 2x dx = x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\)∫tan 2x dx
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\) log|sec 2x| + C
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{4}\) log|sec 2x| + C

Question 10.
∫x cot² x dx on I ⊂ R\{nπ : n ∈ Z}.
Solution:
∫x cot² x dx
= ∫x (cosec² x – 1)dx
= ∫x cosec²x dx – ∫x dx
= x (- cot x) + ∫cot x dx – \(\frac{x^2}{2}\)
= – x cot x + log|sin x| – \(\frac{x^2}{2}\) + C

Question 11.
∫ex(tan x + sec² x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
f(x) = tan x ⇒ f'(x) = sec² x dx
I = ∫ex[f(x) + f'(x)] dx = ex. f(x) + C
= ex. tan x + C

Question 12.
∫ex(\(\frac{1+log x}{x}\))dx on (0, ∞).
Solution:
∫ex(\(\frac{1+log x}{x}\))dx = ∫ex(log x + \(\frac{1}{x}\))dx
= ex.log x + C

Question 13.
∫eaxsin bx dx on R, a, b ∈ R.
Solution:
Let I = ∫eaxsin bx dx …………. (1)
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 3

Question 14.
∫\(\frac{x.e^2}{(x+1)^2}\) dx on I ⊂ R\{-1}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 4

Question 15.
∫\(\frac{dx}{(x^2+a^2)^2}\), (a > 0) on R.
Solution:
Put x = tan t
Then dx = a sec² t dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 5

Question 16.
∫ex log (e2x + 5ex + 6) dx on R.
Solution:
∫ex log (e2x + 5ex + 6) dx
∵ e2x + 5ex + 6 = (ex + 2)(ex + 3)
= ∫ex. log((ex + 2)(ex + 3))dx
= ∫ex.{log(ex + 2) + log(ex + 3)}dx
= ∫exlog (ex + 2)dx + ∫exlog (ex + 3)dx
Put ex = t ⇒ exdx = dt
= ∫log(t + 2)dt + ∫log(t + 3) dt
= log(t + 2)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 2). ∫1dt}dt + log(t + 3)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 3). ∫1dt}dt
= t log(t + 2) – ∫\(\frac{1}{t + 2}\))t dt + t log(t+ 3) – ∫\(\frac{1}{t + 3}\)). t dt
= t {log(t + 2) + log (t + 3)} – ∫\(\frac{t}{t + 2}\)dt – ∫\(\frac{t}{t + 3}\)dt
= t log (t² + 5t + 6) – ∫(\(\frac{t+2-2}{t + 2}\))dt – ∫(\(\frac{t+3-3}{t + 3}\))dt
= t. log (t² + 5t + 6) – ∫{1 – \(\frac{2}{t + 2}\)}dt – ∫{1 – \(\frac{3}{t + 3}\)}dt
= t log (t² + 5t + 6) – t + 2 log|t + 2| – t + 3 log|t + 3| + C
= t log (t² + 5t + 6) – 2t + 2 log|t + 2| + 3log|t + 3| + C
= ex log(e2x + 5ex + 6) – 2ex + 2log ex + 2 + 3 log ex + 3 + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 17.
∫ex.\(\frac{x+2}{(x+3)^2}\) dx on I ⊂ R\ {-3}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 6

Question 18.
∫cos (log x) dx on (0, ∞).
Solution:
Put log x = t
x = et
dx = et . dt
I = ∫et . cos t . dt
= et sin t – ∫sin.et dt
= et . sin t + cos t . et – ∫et . cos t dt
2I = et . (sin t + cos t)
I = \(\frac{e^t}{2}\)(sin t + cos t)
= \(\frac{x}{2}\)[sin(log x) + cos(log x)] + C

III. Evaluate the following integrals.

Question 1.
∫x tan-1 x dx, x ∈ R
Solution:
∫x tan-1 x dx = (tan-1 x)\(\frac{x^2}{2}\) – \(\frac{1}{2}\)∫x².\(\frac{1}{1+x^2}\)dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 7

Question 2.
∫x² tan-1 x dx, x ∈ R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 8

Question 3.
∫\(\frac{tan^{-1}x}{x^2}\) dx, x ∈ I ⊂ R\{0}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 9

Question 4.
∫x cos-1x dx, x ∈ (-1, 1).
Solution:
∫x cos-1x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 10
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 11

Question 5.
∫x² sin-1x dx, x ∈ (-1, 1).
Solution:
∫x² sin-1x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 12

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫x log(1 + x) dx, x ∈ (-1, ∞).
Solution:
∫x log(1 + x) dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 13
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 14

Question 7.
∫sin √x dx, on (0, ∞).
Solution:
x = t² ⇒ dx = 2t dt
∫sin √x dx = 2∫t. sin t dt
= 2(t(-cos t) + ∫cos t dt)
= -2t cos t + 2 sin t
= -2√x cos √x + 2 sin √x + C

Question 8.
∫eax sin (bx + c)dx, (a, b, c ∈ R, b ≠ 0) on R.
Solution:
Let I = ∫eax sin (bx + c)dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 15
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 16

Question 9.
∫ax cos 2x dx on R (a > 0 and a ≠ 0).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 17

Question 10.
∫tan-1(\(\frac{3x-x^3}{1-3x^2}\)) dx on I ⊂ R\{-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)}.
Solution:
Put x = tan t ⇒ dx = sec² t dt
Then
∫tan-1(\(\frac{3x-x^3}{1-3x^2}\)) dx
= ∫tan-1(\(\frac{3 \tan t -\tan^3 t}{1-3\tan^2 t}\)) dx
= ∫tan-1(tan 3t).sec² t dt
= 3∫t sec² t dt
= 3[t∫sec² t dt – ∫{\(\frac{d}{dt}\)(t) ∫sec² t dt}dt]
= 3[t(tan t) – ∫(1) tan t dt]
= 3(x tan-1 x – log\(\sqrt{1+x^2}\)) + C
= 3x[tan-1 x – \(\frac{3}{2}\)log (1 + x²) + C
= 3x tan-1(x) – \(\frac{3}{2}\)log (1 + x²) + C

Question 11.
∫sinh-1 x dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 18

Question 12.
∫cosh-1 x dx on (1, ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 19
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 20

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 13.
∫tanh-1x dx on (-1, 1).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 21