Prasanna – Page 251

RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24A

January 31, 2023

RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A.

Other Exercises

Question 1.
Solution:
(i) When a coin is tossed, we get outcomes 2 as H or T (Head or Tail)
(ii) When two coins are tossed together, we get possible four outcomes as HH, HT, TH, TT
(iii) A die is thrown, we get possible outcomes as 1,2, 3, 4, 5, 6
(iv) From a well – shuffled deck of 52 cards, 0ne card is at random drawn, we get the possible outcomes is 52

Question 2.
Solution:
Possible outcomes = 2
In a single throw of a coin, we get
probability of getting a tail = \(\\ \frac { 1 }{ 2 } \)

Question 3.
Solution:
In a single throw of two coins, possible outcomes = 4
(i) Probability of getting both tails = \(\\ \frac { 1 }{ 4 } \)
(ii) Probability of getting at least one tail = \(\\ \frac { 3 }{ 4 } \)
(iii) Probability of getting at the most one tail = \(\\ \frac { 2 }{ 4 } \) = \(\\ \frac { 1 }{ 2 } \)

Question 4.
Solution:
In a bag, there are 4 white and 5 blue balls ,
Possible outcomes = 4 + 5 = 9
One ball is drawn at random, then
(i) the probability of a white ball = \(\\ \frac { 4 }{ 9 } \)
(ii) the probability of a blue ball = \(\\ \frac { 5 }{ 9 } \)

Question 5.
Solution:
In a bag, there are 5 white, 6 red and 4 green balls
Possible outcome is 5 + 6 + 4 = 15
One ball is drawn at random, then
(i) Probability of a green ball = \(\\ \frac { 4 }{ 15 } \)
(ii) Probability of a white ball = \(\\ \frac { 5 }{ 15 } \) = \(\\ \frac { 1 }{ 3 } \)
(iii) Probability of a non-red ball = \(\\ \frac { 5+4 }{ 15 } \)
= \(\\ \frac { 9 }{ 15 } \)
= \(\\ \frac { 3 }{ 5 } \)
(5 white and 4 green balls are non-red balls)

Question 6.
Solution:
In a lottery, there are 10 prizes and 20 blanks
Possible outcomes = 10 + 20 = 30
A ticket is chosen at random, then
probability of getting a prize = \(\\ \frac { 10 }{ 30 } \) = \(\\ \frac { 1 }{ 3 } \)

Question 7.
Solution:
In a ,box of 100 electric bulb, 8 are defective
Then non-defective bulbs = 100 – 8 = 92
Now possible outcomes = 100
(i) Probability of a drawn bulb, which is defective = \(\\ \frac { 8 }{ 100 } \) = \(\\ \frac { 2 }{ 25 } \)
(ii) Probability of a drawn bulb which is non defective = \(\\ \frac { 92 }{ 100 } \) = \(\\ \frac { 23 }{ 25 } \)

Question 8.
Solution:
A die is thrown, then
Possible outcomes = 6
(i) Now probability of getting 2 = \(\\ \frac { 1 }{ 6 } \)
(ii) Probability of a number less than 3 (which are 1 and 2) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)
(iii) Probability of a composite number (a composite number is a number which is not a prime number which are 4, 6) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)
(iv) Probability of a number not less than 4 (which are 5, 6) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)

Question 9.
Solution:
Total number of ladies = 200
Those who like coffee = 82
Those who dislike coffee = 118
Possible number of outcomes = 200
One lady is chosen at random, then
(i) Probability of a lady who dislikes coffee = \(\\ \frac { 118 }{ 200 } \)
= \(\\ \frac { 59 }{ 100 } \)

Question 10.
Solution:
19 ball bearing numbers, 1, 2, 3,…19
possible outcomes = 19
A ball is drawn at random from the box, then
(i) Probability of a ball which bears a prime numbers which are 2, 3, 5, 7, 11, 13, 17 and 19 = 8 = \(\\ \frac { 8 }{ 19 } \)
(ii) Probability of a ball which bears an even number which are 2, 4, 6, 8, 10, 12, 14, 16, 18 = 9 = \(\\ \frac { 9 }{ 19 } \)
(iii) Probability of a number which bears a number divisible by 3 which are 3, 6, 9, 12, 15, 18 = 6 = \(\\ \frac { 6 }{ 19 } \)

Question 11.
Solution:
A card’s drawn at random from a deck
of well-shuffled deck of 52 cards Probability = 52
(i) Probability of a card being a king = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(ii) Probability of a card being spade = \(\\ \frac { 13 }{ 52 } \) = \(\\ \frac { 1 }{ 4 } \)
(iii) Probability of a card being a red queen = \(\\ \frac { 2 }{ 52 } \) = \(\\ \frac { 1 }{ 26 } \)
(iv) Probability of a card being a black 8 = \(\\ \frac { 2 }{ 52 } \) = \(\\ \frac { 1 }{ 26 } \)

Question 12.
Solution:
One card is drawn at random from a deck of well shuffled deck of 52 cards
Possible outcomes = 52
(i) Probability of a card being a 4 = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(ii) Probability of a card being a queen = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(iii) Probability of a card being a black card = \(\\ \frac { 26 }{ 52 } \) = \(\\ \frac { 1 }{ 2 } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 23 Pie Charts Ex 23B

January 31, 2023

RS Aggarwal Class 8 Solutions Chapter 23 Pie Charts Ex 23B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23B.

Other Exercises

OBJECTIVE QUESTIONS :
Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Central angles = \(\\ \frac { 250 }{ 2400 } \) x 360°
= \(\\ \frac { 75 }{ 2 } \)
= \(37{ \frac { 1 }{ 2 } }^{ o } \)

Question 2.
Solution:
Answer (c)
Central angle = \(\\ \frac { 35 }{ 100 } \) x 360°
= 126°

Question 3.
Solution:
Answer = (a)
Total number of strength = 1650
Arts stream’s central angle = 48°
No. of students of Arts stream
= \(\\ \frac { 48 }{ 360 } \) x 1650
= 220

Question 4.
Solution:
Answer = (c)
Central angle of students reading novels = 81°
\(\\ \frac { 81 }{ 360 } \) x 100
= \(\\ \frac { 45 }{ 2 } \)
= \(22{ \frac { 1 }{ 2 } } \)%

Hope given RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1

January 31, 2023

RD Sharma Class 8 Solutions Chapter 25 Data Handling III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex 25.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1

Other Exercises

Question 1.
The number of hours, spent by a school boy on different activities in a working day, is given below:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1 1
Present the information in the form of a pie-chart .
Solution:
Total = 24

Now we draw a circle and divide it in the sectors having above central angles as shown in the figure.

Question 2.
Employees of a company have been categorized according to their religions as given below:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1 3
Draw a pie-chart to represent the above information.
Solution:
Total =1080

Now draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 3.
In one day the sales (in rupees) of different items of a baker’s shop are given below :
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1 5
Draw pie-chart representing the above sales.
Solution:
Total = 480

Now draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 4.
The following data shows the expenditure of a person on different items during a month. Represent the data by a pie-chart.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.8
Solution:
Total = 10800

Now we draw a circle and divide it into sector having the above central angles as shown in figure.

Question 5.
The percentages of various categories of workers in a state are given in the following table:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.10
Present the information in the form of a pie-chart.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 6.
The following table shows the expenditure incurred by a publisher in publishing a book:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.12
Present the above data in the form of a pie-chart.
Solution:
Total = 100% = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure

Question 7.
Percentage of the different products of a village in a particular district are given below. Draw a pie-chart representing this information.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.14
Solution:
Total = 100

Now, we draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 8.
Draw a pie-diagram for the following data of expenditure pattern in a family.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.16
Solution:
Total =100

Now we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 9.
Draw a pie-diagram of the areas of continents cf the world given in the following table :
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.19
Solution:
Total = 133.3

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 10.
The following data gives the amount spent of the construction of a house. Draw a pie diagram.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.21
Solution:
Total = 300 (in thousands)

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 11.
The following table shows how a student spends his pocket money during the course of a month. Represent it by a pie-diagram.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.23
Solution:
Total expenditure = 100

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 12.
Represent the following data by a pie-diagram :
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.26
Solution:
1. For family A
Total = 10000

Now, we draw a circle and divide it in sectors having the above central angles as shown in the figure.
(2) For family B
Total = 11680

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 13.
Following data gives the break up of the cost of production of a book:
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.29
Draw a pie – diagram depicting the above information.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors of above central angles as shown in the figure.

Question 14.
Represent the following data with the help of pie-diagram.
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.31
Solution:
Total = 6000 tons

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 15.
Draw a pie-diagram representing the relative frequencies (expressed as percentage) of the eight classes as given below :
12.6,18.2,17.5,20.3,2.8,4.2,9.8,14.7
Solution:
Total =100
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.33
Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure given.

Question 16.
Following is the break up of the expenditure of a family on different items of consumption :
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.34
Draw a a pie – diagram to represent the above data.
Solution:
Total = Rs. 3000

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 17.
Draw a pie-diagram for the following data of the investment pattern in five year plan :
RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.36
Solution:
Total = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Hope given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C

January 31, 2023

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C.

Other Exercises

Give the prime factorization of each of the following numbers.

Question 1.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 1.1
∴12 = 2 x 2 x 3 = 2² x 3.

Question 2.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 2.1
∴18 = 2 x 3 x 3 = 2 x 3²

Question 3.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 3.1
∴ 48 = 2 x 2 x 2 x 2 x 3
= 2⁴ x 3.

Question 4.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 4.1
∴ 56 = 2 x 2 x 2 x 7
= 2³ x 7.

Question 5.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 5.1
∴ 90 = 2 x 3 x 3 x 5
= 2 x 3² x 5.

Question 6.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 6.1
136 = 2 x 2 x 2 x 17
= 2³ x 17.

Question 7.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 7.1
∴ 252 = 2 x 2 x 3 x 3 x 7
= 2² x 3² x 7.

Question 8.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 8.1
∴ 420 = 2 x 2 x 3 x 5 x 7
= 2² x 3 x 5 x 7.

Question 9.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 9.1
∴ 637 = 7 x 7 x 13
= 7² x 13.

Question 10.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 10.1
∴ 945 = 3 x 3 x 3 x 5 x 7
= 3³ x 5 x 7.

Question 11.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 11.1
∴ 1224 = 2 x 2 x 2 x 3 x 3 x 17
= 2³ x 3² x 17.

Question 12.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 12.1
∴ 1323 = 3 x 3 x 3 x 7 x 7
= 3³ x 7²

Question 13.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 13.1
∴ 8712 = 2 x 2 x 2 x 3 x 3 x 11 x 11
= 2³ x 3² x 11².

Question 14.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 14.1
∴9317 = 7 x 11 x 11 x 11
= 7 x 11³

Question 15.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 15.1
∴ 1035 = 3 x 3 x 5 x 23
= 3² x 5 x 23

Question 16.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 16.1
∴ 1197 = 3 x 3 x 7 x 19
= 3² x 7 x 19

Question 17.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 17.1
∴ 4641 = 3 x 7 x 13 x 17.

Question 18.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 18.1
∴ 4335 = 3 x 5 x 17 x 17
= 3 x 5 x 17²

Question 19.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 19.1
∴ 2907 = 3 x 3 x 17 x 19
= 3² x 17 x 19.

Question 20.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C 20.1
∴ 13915 = 5 x 11 x 11 x 23 = 5 x 11² x 23

Hope given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

January 30, 2023

In this chapter 6 Life Processes, students will learn about what is meant by life processes, nutrition, and its forms, human respiration, transportation, an excretory system of human beings and plants.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 6 Life Processes. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 6 – Life Processes solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 6 – Life Processes Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for our free webinar class with best Science tutor in India.

Chapter 6 Life Processes InText Questions

Question 1.
Why is diffusion insufficient to meet the oxygen requirement of multicellular organisms like us ?
Answer:
Every living cell requires oxygen for performing cellular respiration. In unicellular organisms (e.g., Amoeba), the single cell is in direct contact with environment. Oxygen passes into it through diffusion. In simple multicellular organisms (e.g., Hydra), every cell may also get oxygen through diffusion from environment. This is not possible in complex multicellular organisms like humans. The body is covered by dead cells. The living cells are not in contact with external environment. Air containing intercellular spaces are absent. Therefore, quick diffusion cannot occur. Cell to cell diffusion is a very slow process. Passage of oxygen from lungs to toes through cell to cell diffusion will take about three years. Therefore, diffusion cannot meet the oxygen requirement of multicellular organisms like humans.

More Resources

Question 2.
What criteria do we use to decide whether something is alive ?
Answer:
The major criterion which is used to decide whether something is alive is movements. Movements may be that of locomotion (e.g., running of dog), movements of a part (e.g., chewing cud by cow), breathing movements, growth, growth movements (in plants) and movement of molecules in metabolic reactions, maintenance and repair of cellular structures.
Besides movements, other criteria found in living beings that distinguish them from the non-living are presence of protoplasm, self built organisation, self repair, reproduction and various life processes like metabolism, nutrition, respiration, growth, exchange of materials, transportation, excretion and irritability. All living beings have a definite life span and life cycle.

Question 3.
What are outside raw materials used by an organism ?
Answer:
Food by heterotrophic organisms ; carbon dioxide, minerals, sunlight and water by autotrophic organisms ; oxygen by all aerobic organisms.

Question 4.
What processes would you consider essential for maintaining life ?
Answer:
Life processes of nutrition, metabolism, respiration, exchange of materials, transportation, excretion and awareness.

Question 5.
What are the differences between autotrophic nutrition and heterotrophic nutrition ?
Answer:

Autotrophic Nutrition

‘ Heterotrophic Nutrition

1. Food. It is self manufactured.

2. External Energy. An external source of energy is required for synthesis of food.

3. Inorganic Substances. They constitute the raw materials for manufacturing food.

4. Digestion. It is absent.

Food is obtained ready-made from outside.

An external source of energy is not required. The required energy is present in the food obtained from outside.

Inorganic substances are not much required.

An external or internal digestion is required for conversion of complex organic materials into simpler and soluble ones.

Question 6.
Where does the plant get each of the raw materials for photosynthesis ?
Answer:

Carbon Dioxide: At through stomata.
Water: Soil through roots.
Minerals: Soil through roots. ,

Question 7.
What is the role of acid in our stomach ? (CCE 2012, 2013)
Answer:
Hydrochloric acid (HCl) is component of gastric juice. It has five functions,

Softening of food,
Conversion of pepsinogen and prorennin into active forms of pepsin and rennin
Acidify the food for proper action of pepsin,
Killing of microorganisms present in food,
Stoppage of action of salivary amylase.

Question 8.
What is the junction of digestive enzymes ? (CCE 2011)
Answer:
Digestive enzymes are hydrolytic enzymes which bring about hydrolytic splitting of complex organic substances into simple, soluble and absorbable substances, e.g.,

Question 9.
How is small intestine designed to absorb digested food ?
Answer:
Small intestine is lined by epithelium which is specialised to absorb. It has mechanisation to increase its absorbing surface area several times,

Villi: They are transverse folds of intestine wall that not only increase surface area but also reach deep into the lumen of intestine for absorption of digested food. Villi possess blood capillaries and lacteals (lymph vessels) for quick transport of absorbed food,
Microvilli: The columnar cells of the intestinal epithelium have fine microscopic outgrowths called microvilli. Microvilli increase the surface area of epithelial cells.

Question 10.
What advantage does a terrestrial organism possess over aquatic organism with regard to obtaining oxygen for respiration ?
Answer:
Air contains about 21% of oxygen while water haà less than 1% oxygen in dissolved state. A terrestrial organism is able to get several times more oxygen than an aquatic organism.

Question 11.
What are the different ways in which glucose is oxidised to provide energy in various organisms ?
Answer:
NCERT Solutions for Class 10 Science Chapter 6 Life Processes image - 2

Question 12.
How is oxygen and carbon dioxide transported in human beings ? (CBSE AI2008, CCE 2012)
Answer:
Oxygen:

97% in combined state with haemoglobin called oxyhaemoglobin.
3% dissolved in plasma.

Carbon Dioxide:

5-7% as dissolved in plasma.
70% as sodium bicarbonate in plasma.
23% in combined state with haemoglobin called carbaminohaemoglobin.

Question 13.
How are lungs designed in human beings to maximise the area for exchange of gases ?
(CBSE AI 2008, CCE 2011)
Answer:
Each lung has a highly branched respiratory tract called respiratory tree. A primary bronchus divides into secondary bronchi, secondary into segmental bronchi, segmental bronchus into bronchioles which divide into terminal bronchioles, respiratory bronchioles, alveolar sacs and alveoli. Alveoli are small rounded or polyhedral pouches which are extremely thin walled and possess a network of capillaries over their surface. They function as respiratory surfaces. The total area of all the alveoli is more than 80 m². It is several times more than the surface area of the whole human body.

Question 14.
What are the components of the transport system in human beings ? What are the junctions of these components ?
(CCE 2011)
Answer:
Human transport system has two components, blood vascular system and lymphatic system..
Blood Vascular System: It consists of blood, blood vessels (tubes) and heart.
Heart: It is pumping organ of blood vascular system.
Blood is made of plasma and three types of cells — red blood corpuscles, white blood corpuscles and blood * platelets.

Blood Plasma: Transport of nutrients, excretory materials, hormones etc.
1. Antibodies in the form of immunoglobins.
2. Prothrombin and fibrinogen for blood clotting.
Red Blood Corpuscles: Transport of oxygen as oxyhaemoglobin. Transport of about 23% carbon dioxide as carbaminohaemoglobin.
White Blood Corpuscles: Phagocytosis of germ cells, production of antibodies and histamine.
Blood Platelets: Formation of thromboplastin for blood clotting.

Blood Vessels:

Arteries: Taking away blood from heart to different body parts.
Veins: Transporting blood towards heart from various body parts.
Capillaries: Exchange of materials between blood and living cells through tissue fluid.

Lymphatic System: It consists of lymph, lymph vessels and lymph nodes.
Lymph:

Collection of extra tissue fluid and passing it back into blood.
Picking up tissue secretions and passing into blood.
Attracting and carrying germs to lymph nodes.

Lymph Vessel: Collection of lymph and passing the same into veins.
Lymph Nodes:

Lymph organs specialised to filter germs,
Maturation of lymphocytes.

Question 15.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds ? (CCE 2012)
Answer:
Mammals and birds are warm blooded animals. They constantly use energy to maintain their body temperature. They have a higher energy need and require more oxygenated blood for their cells. It is important that their oxygenated blood does not mix up with deoxygenated blood.

Question 16.
What are the components of the transport system in highly organised plants ? (CCE 2010)
Answer:
Transport system of highly organised plants consists of xylem and phloem.
Xylem: It is used in transport of water and minerals. Xylem is made of tracheids, vessels, xylem fibres and xylem parenchyma. Tracheids and vessels constitute the tracheary elements or channels for transport of water and minerals.
Phloem: It is used for transport or translocation of organic solutes or food. Phloem consists of sieve tubes, companion cells, phloem fibres and phloem parenchyma. Sieve tubes constitute the channels for transport of food materials.

Question 17.
How are water and minerals transported in plants ?
Answer:
Water and minerals absorbed by the plant roots are passed into xylem as sap. Sap present in xylem is under tension or negative pressure as mesophyll and other cells of aerials parts lose water to the outside through transpiration.
Development of Negative Pressure: Loss of water by mesophyll and other cells of aerial parts in transpiration increases their suction pressure. They withdraw water from xylem channels. As there are billions of transpiring mesophyll cells withdrawing water from xylem channels, water present in xylem comes under negative pressure of 10-20 atmospheres. However, water column does not break due to

Cohesive force amongst water molecules and
Adhesion force between walls of xylem channels and water moecules.

Rise of Sap (Water and Minerals): Tension or negative pressure of water column results in upward pull just as cold-drink is sucked with the help of straw pipe. Since it develops due to transpiration, it is called transpiration pull. The mechanism of this ascent of sap was put forth by Dixon and Joly in 1894.

Question 18.
How is food transported in plants ?
Answer:
Unlike animals, some materials pass in and out of plants through diffusion. For rapid gaseous diffusion to occur, the plants possess stomata and lenticels. During the daytime the photosynthetic organs obtain carbon dioxide from outside by diffusion. The same is used for synthesis of food. Oxygen is released as a by-product. It passes out of the plant by diffusion. Simultaneously, a lot of water vapours pass out.
Other materials required for building plant body are obtained from soil, e.g., nitrogen, phosphorus, other minerals, water .They are sent to chlorophyll containing organs where food is manufactured. The manufactured food is passed to all parts for utilisation and storage. If the distance between the two is small, the materials reach there by diffusion. If the distance is large, as in most plants, they have to be transported through a proper system of transportation. However, plants have a large proportion of dead cells. They do not move. Therefore, they have low energy needs.
The transport is slow: Plants possess two independent transport pathways having conducting tubes. One is xylem that moves water and minerals from soil to aerial parts. The other is phloem which carries food and hormones from the region of availability (e.g., leaves, storage organs) to the areas of utilisation (all living cells, growing points, storage organs, developing fruits).

Question 19.
Describe the structure and functions of nephron.
Answer:
NCERT Solutions for Class 10 Science Chapter 6 Life Processes image - 3

Question 20.
What are the methods used by plants to get rid of excretory products ?
Answer:
Carbon Dioxide: Land plants obtain it from atmosphere while aquatic plants absorb the same from water. Atmosphere contains over 0-038% carbon dioxide. It enters the leaves through stomata. However, the open stomata also become sites for loss of water in transpiration.
Water: Land plants absorb water from soil through their roots. The absorbed water is transported to photosynthetic areas through xylem. Minerals are also transported alongwith water. They are used in synthesis of different organic substances like sulphur in proteins, phosphorus in nucleic acids, magnesium in chlorophyll, etc. 99% of absorbed water is lost through transpiration. Only a small quantity of water is used in photosynthesis as hydrogen donor. For this, water splits into its components with the help of light energy. The phenomenon is called photolysis of water. Oxygen is evolved.

Question 21.
How is the amount of urine produced regulated ?
Answer:
Amount of urine is regulated by volume of blood and amount of antidiuretic hormone (ADH). Volume of blood is determined by presence or absence of extra water in the body. More blood volume will increase pressure in the glomerulus. It increases the amount of glomerular or nephric filtrate. ADH is not secreted. Dilute urine is allowed to pass through the kidneys. The amount of urine is higher than the normal.
In case the body has no extra water or is deficient of water, lesser glomerular filtrate will be produced. ADH is secreted. It helps in withdrawing a good amount of water from urine. Therefore, only concentrated urine is passed out. Amount of urine is smaller than the normal.

Chapter 6 Life Processes NCERT Chapter End Exercises

Question 1.
The kidneys in human beings are a part of the system for
(a) Nutrition
(b) Respiration
(c) Excretion
(d) Transportation.
Answer:
(c).

Question 2.
The xylem in plants is responsible for
(a) Transport of water
(b) Transport of food
(c) Transport of amino acids
(d) Transport of oxygen.
Answer:
(a).

Question 3.
The autotrophic mode of nutrition requires,
(a) Carbon dioxide and water
(b) Chlorophyll
(c) Sun light
(d) All of the above.
Answer:
(d).

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in
(a) Cytoplasm
(b) Mitochondria
(c) Chloroplast
(d) Nucleus.
Answer:
(b).

Question 5.
How are fats digested in our bodies Where does this process take place ? (CCE 2011)
Answer:
1° infants, fat digestion occurs both in stomach and small intestine. However, fat digestion in stomach is poor as gastric lipase is a weak enzyme. Most of the fat digestion occurs in small intestine (jejunum part). Fat is first emulsified with the help of bile salts. Emulsification converts fat into very fine droplets. They are acted upon by pancreatic and intestinal lipase. Lipase breaks fat into fatty acids and glycerol. The latter are absorbed by villi and passed into their lacterals where fat is again formed.

Question 6.
What is the role of saliva in the digestion of food ? (CCE 2011)
Answer:
Saliva is the secretion of salivary glands that is poured in the buccal cavity for :

Moistening and softening of food for easy crushing by the teeth.
Action of enzyme ptyalin or salivary amylase which converts starch and glycogen of cooked food into sweet sugar maltose.
Conversion of semisolid food into slippery bolus for easy swallowing.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by-products ?
Answer:
Conditions,

Carbon dioxide
Chlorophyll
Sunlight
Water
Proper temperature. By-products. Glucose (product), oxygen (byproduct).

Question 8.
What are the differences between aerobic and anaerobic respiration ? Name some organisms that use the anaerobic mode of respiration.
Answer:

Aerobic Respiration	Anaerobic Respiration
1. Method. It is the common method of respiration.	It occurs permanently only in a few organisms. In others it may occur as a temporary measure to overcome shortage of oxygen.
2. Steps. It is completed in 3 steps—glycolysis, Krebs cycle and terminal oxidation.	There are two steps— glycolysis and anaerobic breakdown of pyruvic acid.
3. Oxygen. It requires oxygen.	Oxygen is not required.

Anaerobic Organisms: Yeast (can also perform aerobic respiration), Lactobacillus (bacterium), Ascaris (Roundworm), Taenia (Tapeworm).

Question 9.
How are alveoli designed to maximise the exchange of gases ?
Answer:
Alveoli are small pouches or sacs. About 300 million alveoli occur inside each lung. The whole surface of an alveolus functions as respiratory surface. As there are about 300 million alveoli in each lung, a very large area of respiratory surface becomes available (about 80 m²) for exchange of gases.

Question 10.
What would be the consequence of deficiency of haemoglobin in our body ?
Answer:
Deficiency of haemoglobin is called anaemia. In anaemia the blood is unable to carry the amount of oxygen required by the body (hypoxia). Lesser energy will be available to the body. The person will feel weak, pale, lethargic and unable to perform vigorous exercise or physical work.

Question 11.
Describe double circulation in human beings. Why is it necessary ?
Answer:
Double circulation is the passage of same blood twice through heart, first from right side to lungs and back to left side for passage to rest of the body to be returned to right side. It consists of two components, pulmonary circulation (from heart to lungs and back) and systemic circulation (from heart to different parts of body and back). In pulmonary circulation deoxygenated blood is converted into oxygenated blood. In systemic circulation oxygenated blood is supplied to all parts of the body. It gets changed into deoxygenated form.
Double circulation ensures supply of oxygenated blood to all body parts for efficient release of energy to ensure higher physical activity and thermoregulation of body. It also provides for direct passage of all deoxygenated blood to lungs for oxygenation.

Question 12.
What are the differences between the transport of materials in xylem and phloem ? (CCE 2014)
Answer:
Differences in Transport of Materials in Xylem and Phloem

Transport in Xylem	Transport in Phloem
1. Components. It consists of water and minerals.	It consists of organic nutrients.
2. Direction. The movement is generally unidirectional.	The movement is multidirectional.
3. Force. It operates through creation of negative tension.	It operates through creation of positive turgor.
4. Active Component. An active component is absent.	An active component is present in the region of loading and unloading.
5. Metabolic Inhibitors. Metabolic inhibitors have no effect on xylem transport.	Metabolic inhibitors inhibit phloem transport.
6. Channels. Tracheary elements are non-living.	Phloem channels are made of living cells.

Question 13.
Compare alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:

Alveoli	Nephrons
1. Shape. They are rounded or polyhedral.	They are elongated tubules.
2. Components. Alveoli are single entities.	Each nephron has two components-Malpighian capsule and renal tubule. Renal tubule has three parts – PCT, loop of Henle and DCT.
3. Blood Capillaries. They are of one type and lie all over the alveoli.	Blood capillaries form two patches—glomerulus and peritubular capillaries.
4. Materials. They deal with respiratory gases.	They deal with body fluids.
5. Function. Alveoli perform exchange of gases between blood and inhaled air.	Nephrons bring about separation of waste products from blood.
6. Processes. Gaseous exchange occurs through diffusion.	Urine formation occurs through ultrafiltration, reabsorption, secretion and osmosis.

PRACTICAL SKILL BASED QUESTIONS

Question 1.
Mention any two functions of epidermis. (CCE 2016)
Answer:

Covering and protecting the internal tissues.
Having stomata for gaseous exchange and transpiration.

Question 2.
Explain why only turgid leaf is selected for the preparation of temporary mount of a leaf peel. (CCE 2016)
Answer:
A peel is removed either by twisting and tearing of a leaf or scraping its lower surface. In both the cases only a turgid leaf can be handled properly. In a wilted or flaccid leaf, a lot of mesophyll remains attached to the surface layer or peel. In a turgid leaf, the compact epidermis is easily separable from loose mesophyll.

Question 3.
In an experiment to prepare temporary stained mount of a leaf epidermal peel, how can extra stain be removed ? What possible outcome would be observed if it is removed with cotton wool ? (CCE 2016)
Answer:
By means of blotting paper which can soak the extra stain from the sides of the cover slip. Use of cotton wool can spread the stain over the slide as well as cover slip.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Get 100 percent accurate NCERT Solutions for Class 10 Science Chapter 6 (Life Processes) explained by expert Science teachers. We provide step by step solutions for the questions given in Class 10 Science textbook as per CBSE Board guidelines from the latest NCERT book for Class 10 Science. The topics and sub-topics in Chapter 6 Life Processes

6.1 WHAT ARE LIFE PROCESSES?
6.2 NUTRITION
6.2.1 Autotrophic Nutrition
6.2.2 Heterotrophic Nutrition
6.2.3 How do Organisms obtain their Nutrition?
6.2.4 Nutrition in Human Beings
6.3 RESPIRATION
6.4 TRANSPORTATION
6.4.1 Transportation in Human Beings
6.4.2 Transportation in Plants
6.5 EXCRETION
6.5.1 Excretion in Human Beings
6.5.2 Excretion in Plants.

We cover all exercises in the chapter given below:-

EXERCISE 6.1 – 4 Questions with Solutions
EXERCISE 6.2 – 5 Questions with Solutions
EXERCISE 6.3 – 4 Questions with Solutions
EXERCISE 6.4 – 5 Questions with Solutions
EXERCISE 6.5 – 3 Questions with Solutions
EXERCISE 6.6 – 13 Questions with Solutions.

Download the free PDF of Chapter 6 Life Processes or save the solution images and take the print out to keep it handy for your exam preparation.

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RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

January 30, 2023

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
In a sphere, the number of faces is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Number of faces of a sphere is 1 (a)

Question 2.
The total surface area of a hemisphere of radius r is
(a) πr²
(b) 2πr²
(c) 3πr²
(d) 4πr²
Solution:
Total surface area of a hemisphere is 37πr² (c)

Question 3.
The ratio of the total surface area of a sphere and a hemisphere of same radius is
(a) 2 : 1
(b) 3 : 2
(c) 4 : 1
(d) 4 : 3
Solution:
Total surface area of a sphere = 4πr²
and total surface area of a hemisphere = 3m²
∴ Ratio 4πr²: 3πr²
= 4 : 3 (d)

Question 4.
A sphere and a cube are of the same height. The ratio of their volumes is
(a) 3 :4
(b) 21 : 11
(c) 4 : 3
(d) 11 : 21
Solution:
Let r be the height of a sphere and cube
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 4.1

Question 5.
The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be
(a) 27π cm³
(b) 36π cm³
(c) 108π cm³
(d) 12π cm³
Solution:
Side of cube = 6 cm
∴ Diameter of sphere cut off from it = 6 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 5.1

Question 6.
A cylmderical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be
(a) 4
(b) 3
(c) 6
(d) 8
Solution:
Let r be the radius of a cylindrical rod = r
Then its height (h) = 8r
Volume = πr²h = πr² x 8r = 8πr³
Radius of spherical ball = r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 6.1

Question 7.
If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Solution:
Let r₁ and r₂ be the radius of two spheres
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 7.1

Question 8.
If the surface area of a sphere is 144π m² then its volume (in. m3) is
(a) 288π
(b) 316π
(c) 300π
(d) 188π
Solution:
Surface area of a sphere = 144π m²
Let r be the radius, then
4πr² = 144π
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 8.1

Question 9.
If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq. cm) is
(a) 100π
(b) 75π
(c) 60π
(d) 50π
Solution:
Radius of a sphere (r) = 10 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 9.1

Question 10.
If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is
(a) π : 2
(b) π : 3
(c) π : 4
(d) π : 6
Solution:
Let side of a cube = a
Then volume of cube = a³
The diameter of inscribed sphere = a
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 10.1

Question 11.
If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is
(a) 2r
(b) 3r
(c) r
(d) 4r
Solution:
Radius of a sphere = r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 11.1

Question 12.
A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 12.1
Solution:
Radius of sphere = r

Question 13.
The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is
(a) 2 : 1
(b) 1 : 1
(c) 2 : 3
(d) 1 : 2
Solution:
Let r be the radius of the sphere, then 4
Volume = \(\frac { 4 }{ 3 }\)πr³
Diameter of circumscribed cylinder = 2r
∴ Radius = r
and height (h) = 2r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 13.1

Question 14.
A cone and a hemisphere have equal bases and equal volumes the ratio of their heights is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) \(\sqrt { 2 } \) : 1
Solution:
Let radius of hemisphere and a cone be r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 14.1

Question 15.
A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1
Solution:
∵ Bases of a cone, hemisphere and a cylinder are same
Let radius of each = r
and height of each = r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere MCQS 15.1

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RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22.

Question 1.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph draw one horizontal line OX and other vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the name of subjects taken at uniform gaps.
(iii) Choose the scale = 1 small division = 1 mark
(vi) Then the heights of various bars will be drawn as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 1.1

Question 2.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis write the years taken on uniform gaps.
(iii) Choose scale : 1 small division = 20 students
(iv) Then the heights of various bars will be drawn as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 2.1

Question 3.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis, write the names of sports taken on uniform gaps.
(iii) Choose the scale : 1 small division = 1 student
(iv) Then the heights of various bars will be drawn as shown on the graph
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 3.1

Question 4.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write cities with a uniform gaps.
(iii) Choose the scale : 1 small division = 200 km
(iv) Then we shall draw the heights of various bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 4.1

Question 5.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write countries
(iii) Choose the scale : 1 small division = 10 year
(iv) Then we shall draw the heights of various bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 5.1

Question 6.
Solution:
We can draw a bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write modes of transport with uniform gaps.
(iii) Choose the scale : 1 small division = 100 Students
(iv) Then we shall draw the heights of bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 6.1

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, number of motorcycles.
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to give data as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 7.1

Question 8.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the names of States given at uniform gaps.
(iii) Choose scale : 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 8.1

Question 9.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis
and y-axis respectively.
(ii) Along x-axis, write the names of animals given at uniform gaps.
(iii) Choose scale : 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 9.1

Question 10.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis write the years taken on uniform gaps.
(iii) Choose scale : 1 small division = 20 export earnings
(iv) Then the heights of various bars will be drawn as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 10.1

Question 11.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the names of years given at uniform gaps.
(iii) Choose scale 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 11.1

Question 12.
Solution:
(i) The bar graph shows the number of members in each of the 100 families of a village.
(ii) 90
(iii) 65
(iv) 5

Question 13.
Solution:
(i) The given bar graph shows the marks obtained by a student in an examination in each of the five subjec ts.
(ii) English.
(iii) From the given graph,
(iv) Mathematics.
RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 13.1

Question 14.
Solution:
(i) Mount Everest is the heighest peak and its heights is 8800 m.
(ii) Highest peak is Mount Everest and lowest peak is Annapurna and their heights are 8800 m and 6000 m respectively.
Ratio = 8800 : 6000 => 22 : 15
(iii) Heights of peaks in ascending order is 6000 m, 7500 m, 8000 m, 8200 m and 8800 m.
(iv) Kanchenjunga peak differ by 600 meter from Mount Everest.

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RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

January 30, 2023

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS

Other Exercises

Question 1.
Find the surface area of a sphere of radius 14 cm.
Solution:
Radius of a sphere (r) = 14 cm
∴ Surface area = 4πr² = 4 x \(\frac { 22 }{ 7 }\) x 14 x 14 cm²
= 2464 cm³

Question 2.
Find the total surface afea of a hemisphere of radius 10 cm.
Solution:
Radius of hemisphere (r) = 10 cm
∴ Total surface area = 3πr²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 2.1

Question 3.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 3.1

Question 4.
The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motor cyclist for riding.
Solution:
Diameter of hollow sphere = 7 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 4.1

Question 5.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 5.1

Question 6.
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Solution:
Edge of a solid cube = 44 cm
∴ Volume = a² = (44)² cm²
= 44 × 44 × 44 cm³
Diameter of a spherical bullet = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 6.1

Question 7.
If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.
Solution:
Radius of a sphere (R) = 2r
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 7.1

Question 8.
If a hollow sphere of intefnal and external diameters 4 cm and 8 cm respectively melted into a cone of base diameter 8 cm, then find the height of the cone.
Solution:
Internal diameter of a hollow sphere = 4cm
∴ Internal radius = \(\frac { 4 }{ 2 }\) = 2 cm
Similarly the outer radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm
∴ Volume of melted used in hollow sphere
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 8.1

Question 9.
The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height of the cone.
Solution:
Radius of a sphere (r) = 5 cm
∴ Surface area = 4πr²
= 4π x 5 x 5 = 100π cm²
Radius of cone (r₁) = 4 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 9.1

Question 10.
If a sphere is inscribed in a cube, find the ratio of the volume of cube to the volume of the sphere.
Solution:
Let edge of a cube = a
Then its volume = a³
∵ A sphere is inscribed in the cube
∴ Diameter of sphere = a
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere VSAQS 10.1

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RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21B.

Other Exercises

Question 1.
Solution:
Frequency distribution table is given below:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 1.1

Question 2.
Solution:
Arranging the given data in increasing order:
312, 324, 356, 365, 378, 400, 435, 472, 506, 548, 565, 570, 584, 596, 617, 630, 674, 685, 700, 736, 745, 754, 763, 776, 780.
Now frequency distribution table is given below :
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 2.1

Question 3.
Solution:
Frequency Distribution table is given below:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 3.1

Question 4.
Solution:
Frequency distribution table is given below
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 4.1

Question 5.
Solution:
Frequency table is given below :
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 5.1

Question 6.
Solution:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B 6.1

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NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

January 30, 2023

NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation

MULTIPLE CHOICE QUESTIONS

Question 1.
An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11 and 3/7 parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct ?
(a) d₁ > d₂> d₃
(b) d₁ > d₂< d₃
(c) d₁< d₂> d₃
(d) d₁ < d₂< d₃
Answer:
(d). Upthrust due to liquid on an object is directly proportional to the density of the liquid.

More Resources

Question 2.
An obj ect weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12 N.
Answer:
(a). Explanation : Weight of liquid displaced by an object = Weight of object in air – weight in liquid.

Question 3.
A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base
(b) maximum when breadth and width form the base
(c) maximum when width and length form the base
(d) the same in all the above three cases.
Answer:
(b). Explanation :

SHORT ANSWER QUESTIONS

Question 4.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force.
If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cubeas compared to the first case for water. Give reason for each case. (CBSE 2012)
(b) A ball weighing 4 kg of density 4000 kg m^-3 is completely immersed in water of density 10³ kg m^-3. Find the force of buoyancy on it. (Given g = 10 ms^-2.)
Answer:
(a) Buoyant force = Vρg. Since, density (ρ) of saturated salt solution is more than that of water. So, the cube will experience greater buoyant force in saturated salt solution. When size of cube is reduced, its volume (V) also reduces. Hence, it will experience less buoyant force then in first case.
NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation image - 2

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 16 Floatation are helpful to complete your science homework.

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RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A

January 30, 2023

RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A 1.1

Question 2.
Solution:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A 2.1

Question 3.
Solution:
RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A 3.1

Hope given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.