Bhagya

The detailed Kaveri Class 9 English Book Solutions and Believe in Yourself Poem Class 9 Question Answer serve as excellent models for writing neat exam answers.

Believe in Yourself Class 9 Question Answer

Believe in Yourself Class 9 Questions and Answers

Reflect and Respond (Page 245)

I. Imagine that you are the person in the image.

Question 1.
What emotions do you feel standing at the base of a difficult task?
Answer:
Standing at the base of a difficult task, I would feel nervous but also excited. There might be a little fear of failure, yet I would also feel determined to try.

Question 2.
What might make you take the first step?
Answer:
Self-belief and a strong desire for personal growth are what finally drive me to take the first step.

II. Think about a time when you had to face a challenge.

Question 1.
What was it, and how did you feel at the start of the journey?
Answer:
A challenge I faced was preparing for an important exam. At the start, I felt anxious and unsure, but I knew it was important for my future.

Question 2.
How did you feel once you made the decision to move forward?
Answer:
Once I decided to move forward, I felt more confident and focused. Taking action reduced my fear and made me believe that I could succeed.

III. What does the phrase ‘believe in yourself’ mean to you? Write some words or phrases you associate with believing in yourself.
Answer:
Believing in yourself means trusting your abilities and having confidence that you can achieve your goals.
Words/Phrases associated with believing in yourself:

• confidence
• self-trust
• courage
• determination
• positive thinking
• inner strength
• perseverance

IV. Select the correct meaning of ‘status quo’ based on the given sentence.
Even though some kids wanted to try new activities, most of them were happy with the status quo and didn’t want any changes.
1. A plan to make things more exciting.
2. A situation to keep things the same.
3. A decision where everything is completely different.
4. A choice to change things quickly without thinking.
Answer:
2. A situation to keep things the same.

Check your Understanding (Pages 247-249)

I. Based on your understanding of the poem, select the correct central idea for each stanza from the options given.

Stanza 1
(i) Facing challenges requires personal responsibility and a clear focus on one’s future.
(ii) Facing challenges is a journey best taken with support and guidance from others.
Answer:
(i) Facing challenges requires personal responsibility and a clear focus on one’s future.

Stanza 2
(i) Fear and uncertainty make it difficult to make choices as the future approaches.
(ii) Facing the future with confidence makes choices more straightforward.
Answer:
(i) Fear and uncertainty make it difficult to make choices as the future approaches.

Stanza 3
(i) Personal growth involves finding balance between comfort and change.
(ii) Personal growth requires leaving behind comfort and embracing change.
Answer:
(ii) Personal growth requires leaving behind comfort and embracing change.

Stanza 4
(i) The first step towards change may feel easy if you place your trust in others to guide you and provide support.
(ii) The first step towards change is difficult, but having self-belief and confidence helps you stay on track.
Answer:
(ii) The first step towards change is difficult, but having self-belief and confidence helps you stay on track.

II. Rhyme Scheme

Fill in the blank to complete the following sentence.

The poem follows a simple, yet effective rhyme scheme ______ that flows steadily through each stanza.
Answer:
abcb

III. Tone

State whether the following statements are true or false.

1. The overall tone of the poem is motivational and encouraging.
Answer: True

2. The tone shifts from thoughtful in the beginning to one of determination by the end of the poem.
Answer: True

IV. Speaker

Fill in the blanks with the correct options from those given in the brackets.

The speaker in this poem is not distant; rather, he/ she comes across as a _______ (stranger/guide)
Answer:
guide

who understands the struggle and is encouraging the reader to take _______ (interest in/control of) his/her own future.
Answer:
control of

The use of direct address ‘You’ creates a close connection, as though the speaker is _______ (talking directly to the reader/addressing the reader from a distance)
Answer:
talking directly to the reader

V. Imagery

Match the phrases from the poem in Column 1 with the imagery they represent in Column 2. An extra representation is given.

Answer:

Column 1	Column 2
1. There is no crowd to see…	(ii) Suggests a solitary journey, stressing individual effort.
2. push you back in fear?	(iii) Evokes the mental barrier that prevents growth.

VI. Symbolism

Select the words/phrases from the box below to complete the given sentences.

1. Comfort and the status quo represent _______ and fear of change, symbolising the comfort zone that holds one back.
Answer:
stagnation

2. The future symbolises the _______ the potential for change and success that lies ahead but requires _______ to step into.
Answer:
unknown, courage

3. The first step symbolises the initial _______ required to begin the journey of _______ or personal development.
Answer:
leap of faith, self-improvement

VII. Metaphor The poet uses a metaphor in the line, ‘The first step is the hardest’. Explain why this is metaphorical.

Answer:
The line ‘The first step is the hardest’ is metaphorical because it does not refer to a literal physical step. Instead, it represents the decision to begin something new, such as facing challenges, making changes, or pursuing goals. The poet uses this metaphor to show that starting a journey often requires courage and determination, making it feel more difficult than the steps that follow. –

VIII. Identify the lines from the poem that show antithesis and explain why it is so.

Answer:
The lines from the poem that show antithesis are:

“Will it pull you forward
Or push you back in fear?”

These lines contrast two opposite ideas – being pulled forward (progress, success, confidence) and being pushed back (fear, hesitation, failure). The parallel structure highlights the choice a person must make about their future, making the message clearer and more powerful.
Another example is:

“There is such ease in comfort
To maintain the status quo,
But this isn’t what we are made for
This isn’t how we grow.”

Here, the poet contrasts comfort and staying the same with growth and progress. This antithesis emphasizes that real development happens only when we move beyond our comfort zone.

Class 9 English Believe in Yourself Question Answer

Critical Reflection (Pages 250-251)

I. Read the extract given below and answer the questions that follow.

1. Step up to the challenge
There is no crowd to see,
It’s just you and the future
And where you want to be.

(i) What does the line, ‘There is no crowd to see’ suggest about facing challenges?
Answer:
The line ‘There is no crowd to see’ suggests that facing a challenge is a personal task that you must do for yourself, even when no one is watching or cheering for you.

(ii) Complete the following suitably.
The line ‘It’s just you and the future’ suggests that ________.
Answer:
a person is responsible for shaping their own future through their choices and actions.

(iii) Fill in the blank with the appropriate word/ phrase from the extract.
Latha will her efforts to improve her vocal performance by practicing harder each day.
Answer:
step up to

(iv) Select the most suitable title for the extract.
(a) The Struggles of Change
(b) Facing the Future Alone
(c) A Journey of Growth
(d) The Power or Fear
Answer:
(c) A Journey of Growth

(v) Complete the analogy by using a word from the extract.
achieve: goal:: face: ________.
Answer:
challenge

II. Answer the following questions.

Question 1.
What is the significance of the metaphor, ‘The first step is the hardest’ in the context of personal growth?
Answer:
The metaphor ‘The first step is the hardest’ highlights that beginning something new often requires courage and determination. It signifies overcoming fear, doubt, and hesitation, which are major barriers to personal growth. Once the first step is taken, the path ahead becomes easier to follow.

Question 2.
What message does the antithesis in the poem convey about the nature of personal development?
Answer:
The antithesis in the poem conveys that personal development involves choosing between opposite paths – comfort or growth, fear or progress. It emphasizes that staying in one’s comfort zone may feel safe, but real improvement happens only when we challenge ourselves.

Question 3.
Do you think the poet’s message is realistic in the context of real-world struggles? (Clue: Evaluate whether simply ‘believing in yourself’ is enough to overcome obstacles or other factors are also necessary.)
Answer:
In the poem, the antithesis in the lines “Will pull you forward” and “Or push you back in fear” reflects that personal development is a constant struggle between two opposite forces. It conveys that growth is not automatic; it is a choice. We must decide whether to be guided by our goals or to let our fears stop our progress.

Question 4.
Consider a situation where you or someone you know had to take a difficult first step towards a goal. How does the poem’s message about the importance of self-belief apply to this situation?
Answer:
A good example could be a student who feels nervous about speaking in front of the class but decides to participate in a debate competition. Taking that first step can be frightening because of the fear of making mistakes or being judged. However, by believing in their abilities, the student gains the confidence to try. The poem’s message applies here because selfbelief encourages a person to overcome hesitation and move forward despite uncertainty. Once the first step is taken, it often leads to improvement, new opportunities, and personal growth. The poem reminds us that trusting ourselves is essential when facing challenges and working toward our goals.

Vocabulary in Context (Page 251-253)

I. The phrase ‘status quo’ is a Latin expression that translates to ‘the state in which’ or ‘the existing state of affairs.’ Over time, it has become a popular term used in English to refer to the current situation or condition, especially when things remain unchanged.

There are other Latin terms commonly used in English Read a few given below along with their meanings.

Now, fill in the blanks in the given sentences with the Latin expressions used in English from the table.
(i) I enjoy reading fantasy books, _______ Harry Potter and Magical Paint Brush.
Answer: e.g.

(ii) After helping Tanya with the homework, Ritu asked for a _______ to borrow her notes next time.
Answer: quid pro quo

(iii) I love all kinds of outdoor activities, such as trekking, hiking, biking, _______ .
Answer: etc.

(iv) The park is not very special _______ ; it becomes more fun when you visit with friends.
Answer: per se

(v) We created an _______ team to organise the school festival.
Answer: ad hoc

(vi) The movie started _______ with the hero already fighting the villain in a huge battle.
Answer: in media res

II. The line, ‘Will it pull you forward/or push you back in fear?’ is a rhetorical question.

1. Read the following rhetorical questions and state what they intend to achieve.

(i) Isn’t it obvious that we must act now? Don’t we all have a responsibility to make a change?
Answer:
These rhetorical questions are meant to urge people to take action and remind them that everyone shares the responsibility to bring about change. They encourage the reader to reflect on their duty instead of remaining passive.

(ii) Will we let fear control us, or will we rise above it?
Answer:
This question pushes readers to reflect on fear and choose courage, inspiring them to face challenges and take positive steps forward.

2. Match the situations in Column 1 to the rhetorical questions in Column 2.

Answer:

Column 1	Column 2
(i) Deciding whether to stand up for what is right	F. How can we stay silent when we know what is right?
(ii) Owning up to a mistake made in a group project	C. If I don’t take responsibility now, when will I?
(iii) Deciding whether to try something challenging, like public speaking	E. What’s the point of playing it safe if it means staying stuck?
(iv) Choosing between two career paths	D. Can I really move forward without knowing which path to take?
(v) Deciding whether to apologise for a mistake	B. Isn’t it better to admit our mistakes than to let them define us?
(vi) Trying something new and stepping out of your comfort zone	A. How can we ever grow if we never try anything new?

Listen and Respond (Page 253)

(Refer to the transcript on Page 275 of the NCERT textbook. Listen to its audio using the QR code and answer the questions.)

I. You will listen to a conversation between two friends. As you listen, answer the following questions in one to three exact words that you hear.

Question 1.
How did the boy feel before the play?
Answer:
Nervous

Question 2.
According to the girl, where does confidence come from?
Answer:
Taking action

Question 3.
What was the girl finally sure about regarding the boy?
Answer:
He’ll shine

II. You will once again listen to the conversation. As you listen, select the four true statements from 1-7 given below.

1. The boy feels confident but is worried about freezing on stage.
2. The girl encourages the boy to push through his doubts by trusting in his preparation.
3. The boy thinks that the audience will be supportive regardless of his performance.
4. The girl believes that self-doubt is a normal part of preparing for a big performance.
5. The boy feels that his preparation is not enough and doubts his abilities, even though he has practiced.
6. The girl suggests that the boy should avoid feeling nervous and perform perfectly.
7. The girl believes that pushing through nervousness will help the boy grow and build confidence.
Answer:
2. The girl encourages the boy to push through his doubts by trusting in his preparation.
4. The girl believes that self-doubt is a normal part of preparing for a big performance.
5. The boy feels that his preparation is not enough and doubts his abilities, even though he has practiced.
7. The girl believes that pushing through nervousness will help the boy grow and build confidence.

Speaking Activity (Pages 254-255)

1. Work in pairs. Read the three Sayings/Proverbs given below. For each one, think of a real-life situation where it could apply. Do the suggested role-play and use the Saying/Proverb.
Remember:

• Before diving into a real-life situation, briefly explain the meaning of the saying.
• Describe the situation in the following way-
  • Introduction: Briefly explain the saying.
  • Situation/Example: Share a personal story or relate the saying to something familiar.
  • Conclusion: Reflect on what the saying teaches or what you learned from the experience.
• For the role-play, work in pairs to create the dialogues first and then enact. Take turns to choose your sayings/ proverbs.

Sayings/Proverbs

1. Don’t judge a book by its cover.

• Situation to consider: Have you ever misjudged someone based on their appearance or first impression? What happened?
• Role-play: One student acts as someone who judges others based on their looks or first impressions. The other student plays someone who shows the true personality after a deeper conversation, proving that appearances can be deceptive.

2. Actions speak louder than words.

• Situation to consider: Think of a time when someone’s actions proved more valuable than their words.
• Role-play: One student talks about how they will improve their grades, and the other shows how they actually put in the effort. Compare the impact of talking versus doing.

3. When the going gets tough, the tough get going.

• Situation to consider: Can you think of a time when you faced a difficult challenge but didn’t give up?
• Role-play: One student plays someone who is giving up on a tough school project, while the other encourages that student to keep going. Show how perseverance leads to success despite difficulties.

Sample Answer:
1. Don’t judge a book by its cover.

• Introduction: This proverb means we should not form opinions about someone based only on their appearance or first impression. People may be very different from how they look.
• Situation /Example: Once, a new student joined our class. He was very quiet and wore simple clothes. Some students thought he wouldn’t be very smart. Later, during a science quiz, he answered the toughest questions correctly and even helped others understand the topic.
• Conclusion: I learned that we should take time to know people before judging them. Appearances can be misleading.
• Role-Play: Student A (Judging): He looks so shy and doesn’t talk much. I don’t think he’ll do well in the debate.
• Student B (Proving true personality): Actually, I’ve spoken to him. He reads a lot and has strong opinions.
• Student A: Really? I didn’t expect that.
• Student B: See? You shouldn’t judge a book by its cover.
• Student A: You’re right. I should get to know him better.

2. Actions speak louder than words.

• Introduction: This proverb means that what we do is more important than what we say. Promises mean little without effort.
• Situation/Example: My friend kept saying he would improve his maths marks. But he never studied seriously. Before the next exam, he finally started practicing daily and asking doubts. His marks improved because of his hard work, not just his words.
• Conclusion: I learned that real change happens only when we act, not when we just talk about it.
• Role-Play: Student A (Talking): I promise I’ll score above 90% this term. I’ll definitely do better.
• Student B (Doing): That’s good, but have you started studying?
• Student A: Not yet, but I will soon.
• Student B: I’ve already made a timetable and finished two chapters. Actions speak louder than words.
• Student A: I understand. I need to start working, not just talking.

3. When the going gets tough, the tough get going.

• Introduction: This proverb means that strong and determined people do not give up when things become difficult. Instead, they work harder.
• Situation/Example: During a group project, our model broke just one day before submission. We felt like giving up. But we stayed after school, rebuilt it, and submitted it on time. Our effort paid off.
• Conclusion: I learned that difficulties are not the end. If we stay determined, we can overcome challenges.
• Role-Play: Student A (Giving up): This project is too hard. I can’t finish it. I’m quitting.
• Student B (Encouraging): Don’t give up. We can divide the work and complete it together.
• Student A: But it’s taking too much time.
• Student B: When the going gets tough, the tough get going. Let’s try once more.
• Student A: Okay. Let’s do it.
• Student B: That’s the spirit!

Writing Task (Pages 255-256)

I. Your class is conducting the morning assembly. You have been asked to deliver a speech on the topic, ‘Turning Challenges into Opportunities’. Draft this speech by following the guidelines given below.

• Opening paragraph: Greet the audience and introduce yourself. Begin with a quotation, a question, or a surprising fact. State the purpose of your speech and provide an overview of what you will be talking about.
• Paragraphs 2 and 3-Body of the speech: Divide the body of your speech into two paragraphs, each focusing on a different main point or idea. Use transition words, anecdotes, statistics, and other supporting evidence to strengthen your points.
• Concluding paragraph: Summarise the main points that of your speech. End with a statement that leaves a lasting impression on the audience and convey your thanks.
• Use formal language to present ideas clearly
• Use a persuasive tone-don’t you agree…/…, isn’t it?

Refer to the guidelines given below.

• Consider how challenges can lead to new learning experiences, growth, or unexpected benefits.
• Why do you think people often feel scared or anxious when faced with change?
• How can we open doors to new opportunities? Can you think of an example from your own life or someone you know?
• What are some strategies or attitudes you can adopt to approach change with a positive mindset?
• Think about how facing challenges builds skills, determination, and confidence, all of which are essential for success.

Answer:
Turning Challenges into Opportunities
Good morning, respected Principal, teachers, and my dear friends. I am Kavya/Karan of Class IX, and today I stand before you to speak on the topic ‘Turning Challenges into Opportunities.’

Helen Keller once said, “Character cannot be developed in ease and quiet. Only through experience of trial and suffering can the soul be strengthened.” Isn’t it true that the moments which test us the most are the ones that shape us the strongest? Today, I would like to share how challenges help us grow, why we often fear them, and how we can transform them into stepping stones for success.

To begin with, challenges push us beyond our comfort zones and help us discover hidden strengths. When we face difficulties, we are forced to learn new skills and think creatively. For instance, during the pandemic, students had to suddenly adapt to online learning. At first, many of us felt confused and anxious. However, gradually, we developed digital skills, time management, and self-discipline. What seemed like a disruption actually became an opportunity to become more independent learners. Similarly, when I once participated in a debate competition, I was extremely nervous. Yet, through preparation and practice, I not only improved my speaking skills but also gained confidence. Therefore, challenges often become the training ground for growth.

Furthermore, people often feel scared when faced with change because uncertainty creates doubt. We fear failure, judgment, and the possibility of making mistakes. But don’t you agree that avoiding challenges limits our potential? Change opens doors to new opportunities if we approach it with a positive mindset. Instead of asking, “Why is this happening to me?” we can ask, “What can this teach me?” By staying determined, setting small goals, and believing in ourselves, we build resilience. Each obstacle we overcome strengthens our confidence and prepares us for bigger responsibilities in life. History shows us that many successful individuals turned failures into foundations for success.

If they had given up, would they have achieved greatness? In conclusion, challenges are not roadblocks but stepping stones. They teach us valuable lessons, build character, and open unexpected paths toward success. When we change our perspective, we transform problems into possibilities. So, the next time you face a difficulty, ask yourself: will I step back in fear, or will I step forward with courage? Let us choose courage, growth, and determination.
Thank you, and have a wonderful day.

Learning Beyond the Text (Page 256-257)

I. Explore the stories of Indian leaders who faced significant challenges and turned them into opportunities for success. (Read about the leaders such as Dr. B. R. Ambedkar, Lal Bahadur Shastri, and Dr. APJ Abdul Kalam in the NCERT Pages 256-257)

Find out about such personalities from your village, city, or state and present their success stories in class.
Answer:
Do it yourself.

Hints:

• Choose a personality from your village, town, city, or state (for example, a teacher, social worker, entrepreneur, sportsperson, or freedom fighter).

Find out:

• What challenges they faced.
• How they overcame those challenges.
• What achievements they accomplished.
• What lesson we can learn from their life.

Read the poem ‘Always Believe in Yourself’ Always Believe in Yourself

Always believe in yourself.
Do not limit yourself.
Be kind to yourself
And always believe in all that is good.
You have all the intelligence and ability that you need.
You can attain whatever you are after.
Even though it may not always come the way you believe it should.
Be ready to achieve your dreams.
Believe in yourself when you’re tested beyond your endurance; continue and persist.
Hold on to courage.
Let laughter and encouragement surround you.
The world has much to give;
Always think big,
And keep your hands and heart open
For then you will receive
All of life’s gifts.

Dorothy Hewitt

The detailed Kaveri Book Class 9 Solutions and Class 9 English Kaveri Chapter 2 The Pot Maker Question Answer serve as excellent models for writing neat exam answers.

The Pot Maker Class 9 Question Answer

The Pot Maker Class 9 Questions and Answers

Reflect and Respond (Page 33)

I. Look at the pictures given below and identify the vocations. Now, list at least five more vocations.

Answer:
The vocations shown in the pictures are basket weaving, pot making, handloom (cloth weaving), and carpentry. Five more vocations are blacksmith, cobbler, painter, sculptor, and tailor.

II. Work in pairs. Discuss the following questions and share your answers with your classmates and teacher.

1. What is common among these pictures?
Answer:
They are all vocational works.

2. We refer to such skill-based work as v …. …. a …… ……. o ……. s
Answer:
vocations

3. Mention a few differences between handmade and machine-made products.
Answer:
Handmade products are created using traditional skills and human effort, making each item unique. They require time and patience. Machine-made products are mass-produced, uniform, quicker to manufacture, and depend on mechanical processes rather than craftsmanship.

Check Your Understanding (Page 37)

I. Do you think pot making is easy? Why/Why not?
Answer:
No, pot making is not easy. It involves several laborious steps such as digging clay, pounding it, shaping pots, drying, and firing them carefully. Each stage requires skill, patience, and precision, and even a small mistake can ruin the entire batch.

II. Would Sentila be able to fulfil her dream of becoming a pot maker? Explain.
Answer:
Yes, Sentila would be able to fulfil her dream. Her determination, constant observation, and willingness to learn helped her master the craft despite repeated failures and discouragement.

III. Do you think Mesoba and Arenla would support Sentila? Give a reason.
Answer:
Mesoba would support Sentila because he understands her deep interest and recognises her natural ability in pot making. He represents a more accepting and understanding attitude towards individual talent. Whereas, Arenla did not support Sentila at first because her own experience made her see pottery as tiring work with little reward. She wanted Sentila to have a more secure and less physically demanding life. After the village council’s objection, she started guiding her but not wholeheartedly. Arenla’s resistance reflects concern for her daughter’s well-being and future earnings rather than a lack of love.

Check Your Understanding (Pages 41 & 42)

I. Do you think Onula’s support helped Sentila? If yes, why? If no, why not?
Answer:
Yes, Onula’s support helped Sentila immensely. She reduced Sentila’s fear, boosted her confidence, and taught her calmly. This emotional encouragement helped Sentila relax and learn effectively.

II. Sentila observed her mother making pots. What does this tell you about her?
Answer:
It shows that Sentila was observant, patient, and eager to learn. She believed learning required attention and practice rather than immediate success.

III. Arrange the following events of the story in the correct sequence. Share your answer with your classmates and teacher.

Answer:

1. The village council called Mesoba to know about Arenla’s unwillingness to teach pottery to Sentila.	6.
2. Arenla made a new batch of pots and asked Sentila to continue the work as she was unwell.	7.
3. Sentila observed how other expert potters crafted beautiful pots.	5.
4. Sentila was passionate about pottery but did not share it with her mother.	1.
5. Onula guided Sentila in the art of pot making.	4.
6. Sentila overheard her mother saying that pot making was a tiring job and that she earned very little from it.	3
7. Onula observed two rows of pots inside the work shed, which she felt was the work of two people.	9.
8. Sentila learnt the art of pot making for a year from her mother, but was unsuccessful.	2.
9. Sentila was able to make pots quickly and skillfully, just one less than her mother’s.	8.

Class 9 English The Pot Maker Question Answer

Critical Reflection (Pages 42-45)

I. Read the extracts given below and answer the questions that follow.

1. Pounding the stubborn clay inside bamboo cylinders to soften it, is also tedious. So many times I’ve dropped the mould out of sheer exhaustion and have had to start all over again. It takes months to bring out a batch of pots after so much labour. And the reward? A few rupees. But if Sentila learns weaving, she can make much more money besides providing enough cloth for the family. Weaving is not messy like pot making and can be done indoors in all seasons. Also, the time spent on weaving one shawl is much less and the return is handsome.

(i) Choose the correct reason for the given assertion.
(A): The effort in making pots is far greater than the returns.
A. The process of pot making is quite tiresome and long, and one hardly earns much.
B. The process of pot making is exhausting but gives a sense of satisfaction.
Answer:
A. The process of pot making is quite tiresome and long, and one hardly earns much.

(ii) Why does Arenla want Sentila to learn weaving?
Answer:
Arenla wants Sentila to learn weaving because it provides better financial returns, requires less time, can be done indoors in all seasons, and helps the family by producing cloth, unlike pot making, which is laborious and poorly paid.

(iii) State one advantage of weaving over pot making, as per the extract.
Answer:
One advantage of weaving is that it gives a handsome return in much less time compared to pot making.

(iv) Choose the sentence that uses the word ‘handsome’ in the same way as in the extract.
A. My father is a handsome man.
B. They will make a handsome profit selling this property.
Answer:
B. They will make a handsome profit selling this property.

(v) ‘And the reward?’ What is the author’s purpose in using a question mark here?
Answer:
The author uses the question mark rhetorically, as no answer is expected from this question. The reward is already understood to be meagre and disappointing, despite the exhaustive hard work of pot-making.

2. Onula saw her taking out some clay and the implements from her basket quietly. She watched Sentila’s clumsy efforts to make a pot and noticed that Sentila was too tense. As a result, the clay seemed unable or unwilling to yield the right shape. When Sentila wearily let the misshapen lump fall flat on the ground, Onula went to her and said, “Don’t worry, little one, I shall teach you how to make a perfect pot.” Sentila watched in amazement as Onula fashioned a beautiful pot and asked her to try again.

(i) Complete the sentence with an appropriate reason.
Onula feels Sentila’s effort at making a pot is clumsy because ________.
Answer:
Sentila is too tense, which prevents the clay from taking the right shape.

(ii) Choose the correct option to complete the following sentence.
‘Don’t worry, little one, I shall teach you how to make a perfect pot’.
This shows that Onula was ________ .
(a) sincere and generous
(b) forgiving and thoughtful
(c) thoughtful and generous
(d) forgiving and sincere
Answer:
(c) thoughtful and generous

(iii) Which among the following is the effect of a cause?
A. As a result, the clay seemed unable or unwilling to yield the right shape.
B. Onula saw her taking out some clay and the implements from her basket quietly.
Answer:
A. As a result, the clay seemed unable or unwilling to yield the right shape.

(iv) ‘Onula fashioned a beautiful pot.’ Here the word ‘fashioned’ means ________ ,
Answer:
created/styled

(v) How might Sentila have felt when she saw ‘the misshapen lump fall flat on the ground’?
Answer:
Sentila might have felt ashamed, frustrated, and discouraged because her repeated efforts failed and the clay did not take the desired form.

II. Answer the following questions.

Question 1.
Describe the process of pot making followed by expert pot makers, as observed by Sentila.
Answer:
Expert pot makers dig clay from the riverbank, soak and pound it inside bamboo cylinders to soften it, shape it on the wheel, dry the pots in the sun, arrange them carefully in the kiln with hay and bamboo, and fire them with controlled heat to avoid damage.

Question 2.
What warning was given to Mesoba by the village council?
Answer:
The warming given to Mesoba by the village council was that he must remind Arenla to teach Sentila the art of pot making. The council cautioned that if Arenla did not pass on her skills, this traditional craft could die out. They emphasised that such skills were not the personal property of any individual but belonged to the community, symbolising its tradition and history. The skill should be handed down from one generation to the next. Therefore, experts like Arenla were obliged to share their knowledge not only with their own children but also with anyone willing to learn.

Question 3.
How did Sentila feel when she failed at pot making even after a year of training with her mother?
Answer:
Sentila felt ashamed, frustrated, and discouraged because despite long practice, she could not learn the skill and saw her mother succeed effortlessly.

Question 4.
‘Onula stood there for a long time as if trying to absorb a new phenomenon’. Explain.
Answer:
Onula stood there in disbelief and wondered after seeing two perfect rows of pots-one made by Sentila’s mother and the other by Sentila-realising that she had finally mastered the skill.

Question 5.
‘The tradition and history of the people did not belong to any individual.’ What does this symbolise?
Answer:
It symbolises that traditional skills are a shared cultural heritage meant to be preserved and passed on for the benefit of the entire community, not owned by one person.

Question 6.
What is the significance of the concluding line of the story, ‘A new pot maker was born’?
Answer:
The line signifies the successful transfer of traditional knowledge and marks Sentila’s transformation into a skilled artisan, ensuring the survival of the craft.

Question 7.
What is the role of perseverance in pursuing one’s dreams? Elaborate with reference to Sentila.
Answer:
Sentila’s perseverance helps her overcome repeated failure, discouragement, and resistance. Despite setbacks, she continues learning, which ultimately leads to her success as a pot maker.

Vocabulary and Structures in Context (Pages 45-48)

I. Read the highlighted words in the following sentences from the text.

1. She taught Sentila how to dig the clay with a dao, load it onto her carrying basket…
2. Sentila was a quick learner and turned the clay into malleable dough. Pounding the stubborn clay inside bamboo cylinders to soften it…
The highlighted words describe the tools and materials required in the process of pot making.
Now, classify the words/phrases given in the box as shown in the table below. One example for each category has been done for you.

Answer:

Tools / Implements	Raw Materials	Processes
dao	dough	pounding
kiln	clay	rotating
bamboo cylinders	bed of hay	shaping
spatula
basket

II. Notice the use of the following words in the text:

These words relate to livelihood and economic aspects that are crucial for any enterprise/business/vocation.

Work in pairs and find the meanings of the following words related to the economy. You may refer to a dictionary.

Now, frame sentences using each word.
Answer:

Word	Meaning
bankrupt	unable to pay debts
fiscal	related to government finance
credit	money given with an agreement to repay
inflation	rise in prices
currency	money in circulation
debt	money owed
interest	extra money paid for borrowing
investment	money put into business for profit

Sentences using each word

• The shopkeeper went bankrupt due to lack of sales.
• The government announced a new fiscal policy.
• Small artisans depend on credit for survival.
• Inflation affects the poor the most.
• Currency differs from country to country.
• He cleared his debt after many years.
• The bank charges interest on loans.
• Pottery requires careful investment of time and effort.

III. Read the following sentences from the text. The main clause has been underlined, and the subordinate clause has been circled.

1. Complete the following sentences with suitable noun clauses.
(i) The elders emphasised that ________.
Answer:
traditional skills must be passed on.

(ii) Mesoba explained why ________ .
Answer:
Arenla hesitated to teach pot making.

(iii) Onula’s promise was that ________ .
Answer:
she would help Sentila learn pot making.

(iv) Sentila observed her mother when she was shaping the mouth of the pot, which ________ .
Answer:
looked perfect

(v) The kiln, where ________ required careful attention to prevent over-or under-firing.
Answer:
pots were fired

2. Read the following sentences from the text. Underline the main clause and circle the subordinate clause.

(i) Arenla took Sentila to the riverbank where the grey and red clay was found.
Answer:

(ii) She started on the next one, and like a sprinter who had suddenly found momentum…
Answer:

(iii) … skills such as pot making, which not only catered to the needs of the people…
Answer:

3. Complete the following sentences with suitable relative clauses.

(i) Sentila, whose ________ practised the craft diligently.
Answer:
determination never faded

(ii) The village council, where ________ sought an explanation for Arenla’s reluctance.
Answer:
elders gathered to discuss important matters

(iii) The potter’s hands, which ________ shaped the clay into beautiful creations.
Answer:
which were rough and hardened from years of work

(iv) Arenla, her mother, wanted her to learn weaving, which ________.
Answer:
was considered easier and less demanding than pottery.

(v) Mesoba went home and discussed the matter with Arenla, who ________
Answer:
listened carefully and expressed her concerns about Sentila’s future.

4. (i) Find more determiners from the text.
Answer:
some, many, each, all, one, two, her, their, these, those

(ii) Fill in the blanks with suitable determiners.
(a) The florist arranged five bouquets for her clients that were displayed in ________ elegant flower shop.
Answer: an

(b) The carpenter crafted ________ unique tables, and ________ became the centrepiece in ________ furniture collection.
Answer: some, one of them, the

(c) ________ of. ________ apprentices in ________ culinary class demonstrated ________ knife skills during the intense cooking session.
Answer: Each, the, the, their

(d) ________ of ________ sculptures were displayed at ________ art exhibition, showcasing ________ diverse artistic skills.
Answer: Several, the, the, their

Listen and Respond (Page 48-49)

(Refer to the transcript on Page 261 of the NCERT textbook. Listen to its audio using the QR code and answer the questions.)

I. You will listen to a man speak about stone statues. As you listen, complete the given paragraph by filling in the blanks with the exact words you listen to.
A statue is carved to create a shape that is 1. ________ Among the many things stone is used for making stone 2. ________ is one of them. India has some of the most 3 . ________ stone sculptures, as is obvious from its many stone monuments across the country.
Answer:
1. three-dimensional
2. sculpture
3. fascinating and mesmerising

II. You will now listen to the man speaking about some steps involved in making stone statues. As you listen, select the six correct steps out of the nine given.

1. carve to remove large unwanted portions of the stone
2. set up the different tools
3. measure the weight and dimensions of the statue
4. leave the statue in water to firm up overnight
5. Refine the creation within the stone
6. Choose the stone
7. begin carving from the centre
8. detach the creation from the stone as the final statue
9. work to bring out the imagined shape
Answer:
1. Choose the stone
2. measure the weight and dimensions of the statue
3 . set up the different tools
4. carve to remove large unwanted portions of the stone
5. Refine the creation within the stone
6. Detach the creation from the stone as the final statue

Speaking Activity (Page 49-50)

Work in pairs and choose two characters from the story: Sentila and one other character [Arenla, Mesoba, Onula, or a village elder]
Prepare to speak from the chosen character’s perspective based on information from the story and the understanding of the intentions of the characters.
Prepare a role-play between Sentila and the chosen character. The conversation between the characters should cover the following points.
Sentila’s desire to learn pot making
the challenges she faces
The advice or perspective the other character offers
Students can use direct quotes from the story and creatively expand on the characters’ thoughts and feelings.

You may use the following sentence prompts.

I feel/felt ________ because….
I wish/wished ________ because…
When you said/did ________ it made me feel/think ________ because
Answer:
Do it yourself

Writing Task (Pages 50-51)

Reflective Writing – Identifying Skills and Passion. Reflective writing encourages introspection and thoughtful exploration of personal experiences, skills, and aspirations. It helps individuals gain deeper insights into themselves and their goals through structured reflection.

1. Follow the steps given below to create a write-up about your skills and passions.
Step 1: Introduction
Reflect on your passions and the skills you currently possess or are developing.
Consider why these activities or interests are meaningful and enjoyable to you.
Step 2: Describing skills
Describe specific activities or practices you engage in to nurture your skills. This could include hobbies, classes (art, music, coding, etc.), workshops, or personal projects.
Step 3: Passion into profession
Identify which of these skills you believe have the potential to turn your passion into a profession. Explain why you think these skills are crucial or advantageous in your chosen field.
Step 4: Examples and reflection
Provide examples or anecdotes that illustrate how your skills and passions complement each other. Reflect on how these experiences have shaped your career aspirations and personal growth.
Step 5: Conclusion
Summarise your reflections. Discuss any insights gained about yourself, your skills, and your career ambitions through this exercise.
Answer:
Role Play: Students should enact a conversation between Sentila and another character (Arenla/Onula /Mesoba), highlighting:

• Sentila’s desire to learn pot making
• The resistance she faces
• Emotional conflict and resolution

Learning Beyond the Text (Pages 51-56)

Refer to NCERT (Pages 51-56) for additional information.

The detailed Kaveri Class 9 English Book Solutions and Gifts of Grace: Honouring Our Vocations Poem Class 9 Question Answer serve as excellent models for writing neat exam answers.

Gifts of Grace: Honouring Our Vocations Class 9 Question Answer

Gifts of Grace: Honouring Our Vocations Poem Class 9 Questions and Answers

Reflect and Respond (Page 57)

I. Given below are four riddles. Read and identify who these people are.

Question 1.
In furrows deep, secrets I sow,
As time passes,
I watch them grow.
Who am I? ……….. .
Answer: Farmer

Question 2.
From wheel to kiln, my skill is born,
Step by step, an art takes form.
Who am I? ……….. .
Answer: Potter

Question 3.
I lay foundations, brick by brick,
To build a house, it’s me you pick.
Who am I? ……….. .
Answer: Mason

Question 4.
I work with pots, pans, and spice,
Creating dishes that taste so nice.
Who am I? …………
Answer: Cook

II. What is the role and relevance of the people you identified in the riddles? How do they contribute to society? Discuss with your classmates and teacher.

Answer:
Here is the role and relevance of the people in society I have identified in the riddles:

1. Farmers are the backbone of society, serving as its primary providers. By cultivating grains and other crops, they ensure a steady food supply chain for everyone. Without farmers, fulfilling one of our most basic needs-food-would be extremely difficult.
2. Potters create useful items such as utensils and pots for storing water and other essentials, as well as decorative pieces from clay. They contribute to society by providing both practical tools and artistic creations that enrich daily life.
3. Masons build homes, schools, and roads that shape our living spaces and fulfill basic needs. They contribute to society by creating strong and innovative infrastructure that supports communities.
4. Cooks prepare delicious food that keeps us active throughout the day. They transform raw grains into meals that are both nutritious and culturally significant, sustaining the health and heritage of society.

Check Your Understanding(Pages 60-61)

I. Based on your understanding of the poem, state whether the following statements are true or false. Also rectify the false statements.

1. The poem highlights the skilled work of a craftsman.
Answer: True

2. The poet shares that musicians express emotions through their instruments.
Answer: True

3. The carpenters in the poem are admired for their skilled work.
Answer: True

4. The electricians in the poem are recognised for their crucial role in lighting up lives.
Answer: True

5. The poem pays homage to shoemakers who manufacture quality footwear.
Answer: True

6. The poem celebrates the patriotism of the people of Bharat.
Answer:
False (Correction: The poem celebrates the dignity of labour and the varied vocations of the people of Bharat, not specifically their patriotism.)

7. The poet feels that each vocation deserves to be respected.
Answer: True

II. Let us appreciate the poem.

1. Rhyme Scheme and Lineation

(i) Does the poem strictly adhere to a rhyme scheme, or is it in free verse?
Answer:
The poem does not strictly follow a rhyme scheme. It is written in free verse.

(ii) What is the impact of the varying length of lines in the poem?
Answer:
The varying length of lines creates a natural, flowing rhythm. It gives the poem a conversational tone and emphasizes different professions individually.

(iii) What is the pattern in the structure of most lines of the poem?
Answer:
Most lines begin with ‘The…’ followed by a description of a particular profession. This listing pattern highlights the diversity of vocations.

2. Speaker

(i) Who appears to be the speaker and what is her/his role here?
Answer:
The speaker appears to be a thoughtful observer (possibly the poet) who listens to and appreciates the hardworking people of Bharat.
The speaker’s role is to celebrate and honour their contributions.

3. Tone and Mood

(i) Fill in the blanks with suitable options from the box given below.

A. The tone is _________ and _________ , depicting a sense of admiration and respect for the artisans and craftspersons.
Answer: celebratory; reverential

B. There is a _________ mood throughout the poem, capturing the vibrancy and richness of cultural traditions and skills.
Answer: joyful

4. Imagery

(i) Select any two descriptions from the poem that evoke visual images.
Answer:
Two descriptions that evoke visual images:

• ‘woven with colours and myriad hues’
• ‘gathering their nets from the shore’

(ii) Fill in the blanks with suitable phrases. The poem includes auditory imagery through mentions of artisans with lutes, _________ and _________ emphasising the sounds associated with each vocation.
Answer:
electrician’s humming; boatmen’s singing while at work

5. Metaphor

(i) Select whether the following sentence is true or false.
The mention of ‘delicious singing’ of the cook is a metaphor because it implies that the quality of the singing is so enjoyable or pleasing that it can be equated to the experience of tasting something delicious.
Answer:
True (The phrase ‘delicious singing’ is a metaphor because it compares the cook’s singing to something pleasant like delicious food.)

6. Personification

(i) Select the line that tells us that the poet personifies vocations by attributing humanlike qualities to them.
Answer:
‘The voice of their vocation is the voice of their identity.’
(This line personifies vocation by giving it a ‘voice,’ a human quality.)

7. Repetition

(i) Why might the poet have begun and ended with the same line ‘I hear Bharat celebrating, the varied vocations I hear!’?
Answer:
The poet begins and ends the poem with the same line to create a sense of thematic unity. This structure reinforces the central message that the entire nation is united through its diverse work or occupation. By repeating the same line, he/she ensures the reader stays focused on the primary idea of national celebration.

8. Alliteration

(i) Identify two examples of alliteration from the poem.
Answer:

1. ‘work with cables and wires’
2. ‘sailing and singing while at work’

9. Symbolism

(i) Each vocation in the poem symbolises something more than just a job. What does it symbolise?
Answer:
Each vocation symbolises dignity of labour, identity, and the unity in diversity of Bharat. Together, they represent how every individual contributes to building and strengthening the nation.

Class 9 English Gifts of Grace: Honouring Our Vocations Question Answer

Critical Reflection(Pages 62-63)

I. Read the extract given below and answer the questions that follow.

The shoemakers affirming the quality of their work, for the feet that walk, dance, run, jump, return home. The delicious singing of the cook, or the rhythm of the designer, mason, each celebrating what belongs to them and to none else,
(i) The poet says that the shoemakers ‘affirm’ the quality of their work. What does ‘affirm’ refer to here?
(a) to make adjustments in
(b) to declare with confidence
(c) to label the goods
(d) to justify the efforts
Answer:
(b) to declare with confidence

(ii) What do quality shoes help with, according to the poet?
Answer:
According to the poet, quality shoes help the feet to walk, dance, run, and jump comfortably and safely.

(iii) What does ‘return home’ symbolise besides the literal act of returning?
Answer:
Besides the literal act of returning, the phrase “return home” symbolises finding rest, safety, and a sense of belonging after a long day of hard work.

(iv) Identify the phrase that tells that every worker’s contribution is distinct.
Answer:
The phrase ‘each celebrating what belongs to them and to none else’ tells that every worker’s contribution is distinct.

(v) Complete the following with suitable words. ….for the feet that walk, dance, run, jump, return home refers to
Answer:
The shoemaker, who crafts quality shoes that delight people with their comfort.

II. Answer the following questions.

Question 1.
Why does the poet say, ‘I hear Bharat celebrating, the varied vocations I hear’?
Answer:
The poet says so because the work of common people is full of energy and pride, which makes the whole nation (Bharat) happy.

Question 2.
What does the electrician ‘humming’ while getting ready for work suggest?
Answer:
The electrician ‘humming’ suggests happiness, enthusiasm, and pride in his work. It shows that he does not see his job as a burden but performs it with interest and dedication.

Question 3.
Explain the significance of the line, ‘The voice of their vocation is the voice of their identity.’
Answer:
This line means that a person’s profession reflects who they are. Their work shapes their identity, values, and self-respect. It emphasises that every vocation is meaningful and contributes to a person’s individuality.

Question 4.
Do you think the role of all the people belonging to different vocations is important in our daily lives? Support your answer with a reason.
Answer:
Yes, every vocation is important because society depends on different workers for food, shelter, electricity, clothing, and other needs. Without their contributions, daily life would not function smoothly.

Question 5.
Why is the poet celebrating all the vocations in the poem? Explain by giving examples from your context.
Answer:
The poet celebrates these vocations because every worker takes pride in their specific craft and contributes to the nation’s growth. In our daily lives, we see this when a local tailor carefully stitches clothes or a street vendor prepares food with skill. These roles are essential for our society to function and deserve respect.

Question 6.
How does the poet use sensory imagery to bring out the beauty of everyday work?
Answer:
The poet uses auditory imagery like ‘humming,’ ‘singing,’ and ‘voice,’ and visual imagery such as ‘woven with colours and myriad hues’ and ‘gathering their nets from the shore.’ These descriptions help readers see and hear the beauty in ordinary work, making it lively and meaningful.

Vocabulary in Context(Pages 63-64)

I. People of different vocations are being described in the poem. Match the vocations given in the box below with the descriptions that follow.

1. A person who studies or grows garden plants: _________.
Answer: horticulturist

2. A trained female community health worker: _________.
Answer: ASHA worker

3. A producer of raw silk: _________ .
Answer: sericulturist

4. A person whose job is making or selling sweets and chocolates: _________.
Answer: confectioner

5. A metalworker who specialises in working with precious metals: _________.
Answer: goldsmith

6. A person who fuses materials together: _________.
Answer: welder

II. Identify the word from Column 2 that is not the synonym of the words given in Column 1.

Answer:

1. countable
2. drawing
3. uniform
4. inedible

Listen and Respond (Pages 64-65)

(Refer to the transcript on Page 262 of the NCERT textbook. Listen to its audio using the QR code and answer the questions.)

I. You will listen to two friends, a boy and a girl discussing the occupations of their parents. As you listen, answer the following question in two to three words only.
Answer: Tool kit

II. You will listen to the girl and the boy once again. As listen, answer the questions by selecting the correct option.

Question 1.
The girl is _________ about taking food for her mother at the factory.
(i) happy
(ii) boastful
(iii) unsure
Answer:
(i) happy

Question 2.
The boy thinks that the job of the girl’s mother carries a lot of _________ .
(i) luck
(ii) risk
(iii) responsibility
Answer:
(iii) responsibility

Question 3.
The girl _________ why the boy is good at science exhibitions.
(i) wonders
(ii) questions
(iii) realises
Answer:
(iii) realises

Question 4.
The girl shares her _________ about joining the weekend discussion on tools.
(i) doubts
(ii) wish
(iii) ideas
Answer:
(i) wish

Speaking Activity(Page 65)

The poet speaks of several occupations in the poem. Create groups of five. Allot the role of any five vocations that the poet talks about. Each student in the group will represent one vocation.
Each student speaks for 1-2 minutes about their vocationwho they are, what they do, where they work, what kind of experiences they have at work, what problems they face and a message for all.

Answer:
Five vocations chosen from the poem: Carpenter, Electrician, Boatman, Shoemaker, Cook.
Each speech is written in first person and can be spoken for 1-2 minutes.

1. Carpenter: Hello everyone!
I am a carpenter. I work with wood and create furniture, doors, windows, and sometimes even beautiful decorative pieces. With careful measurement and mathematical precision, I turn simple wood into something useful and lasting. I usually work in workshops or at construction sites. My work requires patience, skill, and creativity. Sometimes I face challenges like working in extreme heat or lifting heavy materials. But I feel proud when families use the furniture I build every day.
My message to all is- ‘Respect skilled worker. Behind every strong home, there are hardworking hands shaping it with care.’

2. Electrician: Good morning.
I am an electrician. I work with cables and wires to bring light and power into homes, schools, and offices. I install, repair, and maintain electrical systems to ensure everything runs smoothly. My work can be risky because I deal with electricity, so I must be very careful and follow safety rules. Sometimes I work long hours, especially during emergencies.
I feel happy knowing that I help ‘brighten lives.’
My message is- ‘Value the people who work behind the scenes to make your life comfortable and safe.’

3. Boatman: Hello friends!
I am a boatman. I gather my nets from the shore and sail into the sea. I fish to provide food for many families and sometimes carry goods or people across waters.
The sea is beautiful but unpredictable. Storms and rough waves can make my work difficult. Yet, I sing while I work because I love what I do.
My message is- ‘Every profession has challenges, but courage and dedication help us move forward.’

4. Shoemaker: Hello friends!
I am a shoemaker. I design and repair shoes that protect your feet when you walk, run, dance, and travel. I ensure that my work is strong, comfortable, and durable.
My work requires careful stitching and attention to detail. Sometimes people overlook my role, but without good shoes, daily life would be uncomfortable.
My message is- ‘Never consider any job small. Every step you take is supported by someone’s effort.’

5. Cook: Hello everyone!
I am a cook. I prepare meals using different ingredients, spices, and recipes. I work in kitchensat homes, restaurants, and events-bringing joy through food. Cooking requires creativity, cleanliness, and patience. Sometimes I work under pressure, especially during busy hours. But I feel satisfied when people enjoy the food I prepare.
My message is- ‘Food brings people together. Respect those who nourish you with care and dedication.’

Writing Task (Page 66-67)

I. Your school will be organising a ‘Career Mela’ in which students are to be made familiar with the different career options. Complete the poster for the event given below.

A poster is used to announce an event. It has the following features:

• placed in a box
• visually attractive
• letters of different fonts and size
• proportionate spacing
• word limit: 50

Answer:
SAPPHIRE PUBLIC SCHOOL
announces
CAREER MELA
to spread awareness about various careers on 25 February from 9 a.m. to 6 p.m. at School Auditorium, Sapphire Public School

Highlights:

• Details, information and guidance provided for all streams
• Counsellors for all career paths
• Interactive sessions with professionals

Event
Purpose: To guide students about various career options and future opportunities
Date: 25 February 20XX
Time: 9:00 a.m. – 6:00 p.m.
Venue: School Auditorium
Slogan: CHART YOUR FUTURE AT CAREER MELA
ENTRY FREE.
Entry Tickets: Free for all students and parents
Sponsors: Local Educational Institutions and Career Guidance Centres
Issuing Authority: Principal, Sapphire Public School

Learning Beyond the Text (Pages 67-68)

I. For self-reading
II. For self-reading

III.

A Japanese Haiku
Make up your mind, Snail!
You are half inside your house, And halfway out!

Haiku is a traditional Japanese poetry form. A very short poem consisting of 17 syllable verses divided into three lines, with five syllables in the first, seven syllables in the second, and five syllables in the third.
Now, create a Haiku poem and share with your classmates and teacher.
Answer:
Here is a haiku following the 5-7-5 syllable pattern:
Morning sun rises
Golden light warms silent hills
Birdsong fills the air

Start by reviewing these Exploration Class 9 Science Solutions Chapter 12 Patterns in Life Diversity and Classification Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 12 Question Answer

Class 9 Science Ch 12 Patterns in Life Diversity and Classification Question Answer

Patterns in Life Diversity and Classification Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
Meena and Hari observed an animal in their garden. Hari called it an insect while Meena said it was an earthworm. Choose the correct option which confirms that it is an insect.
(i) Bilateral symmetrical body
(ii) Body with jointed legs
(iii) Cylindrical body
(iv) Body with little segmentation
Solution:
(ii) Insects belong to Arthropoda and have jointed legs. Earthworms have cylindrical, segmented bodies with no jointed legs.

Question 2.
Sponges represent one of the simplest animal body plans. Their bodies lack true tissues and organs. Which feature of sponge cells supports its classification under the animal kingdom?
(i) Absence of mitochondria
(ii) Ability to photosynthesise
(iii) Presence of a cell membrane
(iv) Presence of a cell wall
Solution:
(iii) Animal cells have a cell membrane but no cell wall. Sponges are animals; they do not photosynthesise, and they do not have cell walls.

Question 3.
Observe two different animals in your immediate environment. What features help you distinguish between them? How do these features help place them into different groups?
Solution:
Two animals, such as a dog and a butterfly, can be distinguished based on several features. The dog has a backbone, fur, and moves by walking, whereas the butterfly has wings, an exoskeleton, and flies. These features help in classification as the dog is a vertebrate (having a backbone), while the butterfly is an invertebrate belonging to the group Arthropoda. Thus, differences in body structure and movement help place them into different groups.

Question 4.
How would a scientist justify choosing cellular organisation as a more fundamental characteristic for the basis of classification rather than the presence of xylem and phloem?
Solution:
Cellular organisation (prokaryote vs eukaryote, unicellular vs multicellular) applies to all living things, including plants, animals, fungi, and bacteria. But xylem and phloem are found only in some plants. So cellular organisation is more fundamental and universal as a basis for classification.

Question 5.
You find an unlabelled slide of a single-celled organism that has a well-defined nucleus and multiple cilia. Which group would it most likely belong to? Give reasons.
Solution:
It most likely belongs to Kingdom Protista (like Paramecium). It is unicellular; it has a true nucleus (eukaryote), and cilia help it move in water. All these features together point to Kingdom Protista.

Question 6.
How does the diversity of organisms contribute to the balance and stability of an ecosystem?
Solution:
Diversity of organisms ensures that different ecological roles, such as producers, consumers, and decomposers, are fulfilled. Plants produce food and oxygen, animals help in pollination and seed dispersal, and microorganisms decompose dead matter and recycle nutrients. This interdependence maintains ecological balance and increases the stability of ecosystems.

Question 7.
If all unicellular organisms were grouped into a single kingdom, what problems would arise?
Solution:
Unicellular organisms vary greatly as some are prokaryotic while others are eukaryotic. Their modes of nutrition also differ, as some are autotrophic while others are heterotrophic. Grouping them into one kingdom would ignore these fundamental differences. This would make classification meaningless and fail to reflect true evolutionary relationships.

Question 8.
Why are viruses not placed in any of the five kingdoms? Give a reason.
Solution:
Viruses are not truly living. They have no cells, cannot carry out life processes on their own, only become active inside a host cell and have no metabolism. All five kingdoms are made of cells. Since viruses have no cellular organization, they do not fit anywhere in the five-kingdom system.

Question 9.
If you were asked to revise the five-kingdom classification, would you create a separate category for viruses or keep them outside the system? Justify your answer and explain what this indicates about the evolving nature of scientific classification.
Solution:
Yes, many scientists suggest viruses deserve their own category since they have genetic material but are acellular. This shows that classification systems are not perfect or permanent, they keep changing as we learn more. Science is always evolving and updating itself based on new discoveries.

Question 10.
Viruses contain genetic material like living organisms but lack cellular organisation. Which features prevent them from fitting into the five kingdom system? What does this tell us about the limitations of classification systems?
Solution:
Viruses lack cellular organization, their own metabolism and the ability to reproduce independently. This tells us that the five kingdom system has limitations as it was designed for cellular life forms. Viruses challenge our very definition of life and show that classification systems need to be flexible.

Question 11.
Both pteridophytes and bryophytes lack flowers and seeds, yet they are placed in different groups. Explain.
Solution:
Even though both bryophytes and pteridophytes do not have flowers or seeds, they are placed in different groups because of important differences in their body structure. Bryophytes are simpler plants that do not have true roots, stems, or leaves and they completely lack vascular tissues like xylem and phloem, which means they cannot transport water and food efficiently to different parts of the plant.

Pteridophytes, on the other hand, are more advanced because they have proper roots, stems and leaves and they also have well-developed vascular tissues that help them transport water and nutrients throughout the plant body. This makes pteridophytes much better adapted to life on land compared to bryophytes. Both still need water for reproduction, but the presence of vascular tissue and true plant organs is the key reason they are classified separately.

Question 12.
In the classification hierarchy, which group, class or genus has fewer members but more features in common? Explain your answer.
Solution:
In the classification hierarchy, genus has fewer members but more features in common. This is because genus is a lower category than class. Organisms in the same genus are closely related and share many similar characteristics, whereas a class includes a larger number of organisms with fewer similarities. Hence, genus is more specific.

Question 13.
A scientist discovers a new organism with the characteristic features of locomotion and autotrophic nutrition. Which character(s) would help the scientist identify the organism as belonging to Protista according to the five-kingdom classification?
Solution:
Protista includes unicellular eukaryotic organisms, and the presence of both locomotion and autotrophic nutrition is characteristic of organisms like Euglena, which belongs to this kingdom. The unicellular and eukaryotic nature of the organism would confirm its placement in Protista.

Question 14.
A researcher identified a unicellular eukaryotic organism as fungi. What identification key would you suggest according to the five-kingdom classification to keep a unicellular organism in the kingdom Fungi?
Solution:
According to the five-kingdom classification, an organism can be placed in kingdom Fungi if it is eukaryotic, unicellular, and heterotrophic with a cell wall made of chitin. The mode of nutrition must be saprophytic or parasitic, obtaining nutrients by absorption from dead or living organic matter. Yeast is the only unicellular organism placed in Kingdom Fungi based on these characteristics.

Question 15.
During a long-term ecological study, students examined organisms collected from three different environments – a freshwater pond, damp soil near decaying logs and the digestive tract of animals. Instead of naming organisms directly, scientists recorded only structural, cellular and nutritional features as given in the table below.

The students realised that some organisms fit neatly into Whittaker’s five kingdom classification, while others challenged the very basis of this classification.
Based on the case study, answer the following questions
(i) Identify one organism that clearly belongs to the Kingdom Fungi. State one observation that supports your answer.
Solution:
Organism Q belongs to the Kingdom Fungi. It is multicellular with a cell wall, has no chlorophyll, and grows on dead organic matter, which is exactly how fungi like mushrooms work.

(ii) Which organism would be placed in the Kingdom Monera? Mention one characteristic that justifies this placement.
Solution:
Organism P belongs to Kingdom Monera. It is microscopic with no true nucleus (prokaryote) and has a rigid cell covering. These are the defining features of bacteria.

(iii) Organisms Rand Q are both eukaryotic, yet they are placed in different kingdoms. Analyse the criteria that separate them.
Solution:
Organism R belongs to Protista as it is unicellular, has flagella, and can do photosynthesis in light but is heterotrophic in darkness. Organism Q belongs to Fungi as it is multicellular, filamentous, has no chlorophyll, and absorbs food from dead matter. The key differences are level of organisation (unicellular vs multicellular), mode of nutrition, and body structure.

(iv) Explain why organism S cannot be classified using the mode of nutrition alone.
Solution:
Organism S is multicellular with a backbone and shows aquatic respiration during early life, which points to a vertebrate like an amphibian or fish. The mode of nutrition (heterotrophic) is shared by animals, fungi, and some protists. So nutrition alone is not enough. We need other features like presence of backbone, tissue organisation, and reproduction to correctly classify it under Vertebrata.

(v) Organism T does not fit into any of the five kingdoms. Which fundamental characteristic used in classification does it lack and what does this reveal about the limitations of classification systems?
Solution:
Organism T (a virus) lacks cellular organisation as it is a cellular and remains inactive outside a host cell. All five kingdoms are based on cellular life forms. This shows the biggest limitation of the five kingdom system, which is that it cannot classify non-cellular entities like viruses.

(vi) If classification were based only on habitat, which organisms might be incorrectly grouped together? Explain the scientific consequences of such a classification.
Solution:
Fish and whales would be grouped since both live in water, even though whales are mammals. Bats and birds would also be grouped together since both fly in air, even though bats are mammals and birds are a separate class. This would be scientifically wrong because their body organisation, reproduction, and evolutionary history are completely different. Habitat-based classification ignores fundamental biological differences and would give a very misleading picture of how organisms are actually related.

(vii) Imagine scientists discover a new organism that is multicellular, eukaryotic, lacks chlorophyll, and absorbs nutrients from a host externally. Should it be placed under fungi or animalia? Justify your reasoning using classification criteria.
Solution:
It should be placed under Kingdom Fungi because it absorbs nutrients externally from the host, which is a typical fungal feature, and it has no chlorophyll.

Animals ingest food internally and digest it inside their body. They do not absorb nutrients from the outside of a host. The external absorption of nutrients is the deciding factor that places this organism closer to Kingdom Fungi than Animalia.

Class 9 Science Chapter 12 Patterns in Life Diversity and Classification Question Answer (InText)

Think It Over

Question 1.
What do you understand by biodiversity?
Solution:
Biodiversity means the variety of living organisms found on Earth. It includes different types of plants, animals, and microorganisms, as well as the different habitats in which they live.

Question 2.
How does the grouping of organisms help us understand diversity?
Solution:
When we group organisms based on their similarities, it becomes easier to study them. We can understand how they are related to each other and how they evolved.

Question 3.
On what basis are plants and animals classified?
Solution:
They are classified based on things like their body structure, how they get food (autotrophic or heterotrophic), their cell type, how they reproduce, and their genetic similarity.

Question 4.
How does classification help address problems in farming?
Solution:
Classification helps farmers identify useful and harmful insects, choose the right crops, understand which plants resist drought or pests and maintain food security.

Pause and Ponder

Question 1.
If many organisms share common features, could they also share a common ancestry?
Solution:
Yes. When organisms share similar features like body structure or cell type, it usually means they evolved from the same ancestor long ago. The more features they share, more closely related they are.

Question 2.
How can a single-celled organism carry out all its life processes when billions of cells are required in multicellular organisms like us?
Solution:
A single-celled organism performs all life processes within one cell, making it self-sufficient. In multicellular organisms, different cells are specialised to perform different functions. No single cell can carry out all functions alone due to this division of labour. Therefore, billions of coordinated cells are needed to sustain life.

Question 3.
Which plant features reduce their dependence on water but still require moist conditions?
Solution:
Bryophytes have root-like rhizoids that help them absorb moisture from damp surfaces. They do not need to be in water, but they still need moisture around them, especially for reproduction.

Question 4.
Why do taller plants need specialised transport tissues?
Solution:
Taller plants cannot absorb water and food directly through their surface like small plants. They need xylem (to carry water upward) and phloem (to carry food) to transport nutrients to every part of the plant.

Question 5.
How do seeds and fruits affect where and how plants can survive?
Solution:
Seeds and fruits help plants spread to new places. Fruits protect seeds and help in their dispersal by wind, water, or animals. This allows plants to grow in different habitats, increasing their chances of survival and reducing competition with the parent plant.

Question 6.
An earthworm (Annelida) and a beetle (Arthropoda) both have segmented bodies, but the beetle has a hard external skeleton. How does the beetle’s external skeleton help it survive?
Solution:
The beetle’s hard outer covering (exoskeleton) protects it from predators, stops water loss from its body, and supports strong muscles. This is why beetles can survive in dry, hot, and exposed places where earthworms cannot.

Question 7.
Does the term ‘biodiversity’ relate only to the variety of organisms, or does it encompass other elements?
Solution:
Biodiversity is more than just counting species. It also includes a variety of ecosystems like forests, oceans, and deserts, genetic diversity within species, and all the relationships between organisms and their environment.

Question 8.
If you find a new organism in a pond, what features will you observe to classify it and why?
Solution:
We would observe whether it is unicellular or multicellular, whether it has a nucleus, whether it moves, how it gets food and whether it has a cell wall. These features would help place it in the right kingdom.

Question 9.
Why do genetic studies provide deep information about living beings?
Solution:
Genetic studies provide information about the internal organisation and functioning of living beings at the molecular level. They reveal evolutionary relationships between organisms and explain the basis of inherited traits and diseases. Variations studied through genetics account for the diversity observed among living beings. Thus, genetic studies offer deep and precise insight into the biology of all living organisms.

Question 10.
How can changes in climate affect biodiversity?
Solution:
Climate change causes temperature rise, changed rainfall patterns, and habitat loss. Many species cannot adapt fast enough and may go extinct. When one species disappears, others that depend on it also suffer; it is like removing one piece from a puzzle.

Class 9 Science Chapter 2 Question Answer (Activities)

Activity: 1 Comparing and Classifying Organisms

Objective:

• To observe different organisms in an ecosystem.
• To classify organisms based on habitat, activity time, and features.
• To understand that organisms can be grouped using different criteria.

Materials Required:

• Given textbook picture (ecosystem image)
• Pencil/pen
• Observation notebook

Procedure:

1. Carefully observe the given ecosystem image (day and night view).
2. Identify as many organisms (animals/birds/insects) as possible.
3. Note where each organism is seen: Air, Tree, Water, Ground/forest floor
4. Observe when each organism appears active: Day, Night, Both
5. Record your observations in the table.
6. Try grouping organisms based on: Habitat, Activity time, Eating habits

Observation Table: Recording Observations

Organism	Where seen (Air/Tree/Water/Ground)	Active (Day/Night/ Both)	Visible Feature(s)
Owl	Tree	Night	Feathers, big eyes
Tiger	Ground	Day	Stripes, sharp teeth
Monkey	Tree	Day	Tail, limbs
Rabbit	Ground	Day	Long ears
Deer	Ground	Day	Horns
Bat	Air/Tree	Night	Wings
Bird (Eagle)	Air	Day	Feathers, wings
Leopard	Tree/Ground	Night	Spots
Frog	Water/Ground	Both	Moist skin
Fish	Water	Day	Fins, scales

Table: Grouping of Organisms

Grouping Criterion	Organisms in Group	Feature Used
Carnivores	Tiger, Leopard, Eagle	Eating habits
Herbivores	Deer, Rabbit	Eating plants
Nocturnal	Owl, Bat, Leopard	Active at night
Diurnal	Monkey, Deer, Eagle	Active during day
Aquatic	Fish, Frog	Live in water
Arboreal	Monkey, Owl	Live on trees

Conclusion:

• Organisms in an ecosystem show diversity in habitat, behaviour, and features.
• The same organism can belong to multiple groups depending on the criteria used.
• Classification helps in organising and understanding living organisms systematically.

Viva Voce Questions

Question 1.
What is meant by classification of organisms?

Question 2.
Why can the same organism be placed in different groups?

Question 3.
What are nocturnal animals? Give one example.

Question 4.
On what basis can organisms be grouped?

Question 5.
Why is classification important in biology?

Activity 2: Case Study – Pakke Tiger Reserve

Objective:

• To understand biodiversity and species distribution in a forest ecosystem.
• To learn how scientists classify and study organisms.
• To analyse the importance of habitat and interdependence among organisms.

Procedure:

1. Carefully read the case study of Pakke Tiger Reserve.
2. Identify key information such as: Number and types of species, Special organisms (e.g. hornbills), Habitat conditions (trees, fruits, forest regions)
3. Observe how different species depend on: Tree size, Food availability
4. Analyse how scientists study biodiversity using patterns.
5. Answer the given case-based questions based on your understanding.

Case study:

Carefully read the given case study of Pakke Tiger Reserve.

The Pakke Tiger Reserve in Arunachal Pradesh is a forest where scientists have recorded nearly 300 bird species, which is striking given that India as a whole has about 1,300 bird species. Pakke is also known for supporting four species of hornbills — the Rufous-necked Hornbill, the Oriental Pied Hornbill, the Great Hornbill and the Wreathed Hornbill.

These large birds nest only in large, old trees with suitable cavities and feed on specific fruits.

As a result, different hornbill species are found in different parts of the forest depending on tree size and fruit availability.

Studying such patterns allows scientists to ask precise questions about biodiversity, such as

• How are species distributed within a forest? Which plants and animals are closely linked?
• How does classifying the four hornbill species help us understand biodiversity?

Think and discuss the case based on the following questions:

1. How can scientists keep track of so many species?
2. The four hornbills look similar in some ways. What features can help scientists distinguish them from one another?
3. What would happen if the large, old trees disappeared from the forest?

Activity 3: Observing Bacteria and Cyanobacteria

Objective:

• To observe bacteria and cyanobacteria under a microscope.
• To understand that they are unicellular prokaryotes.
• To compare their basic structure and identify similarities.

Materials Required:

• Permanent slides of bacteria and cyanobacteria
• Microscope
• Notebook and pencil

Procedure:

1. Take the prepared slides of bacteria and cyanobacteria.
2. Place the slide on the microscope stage.
3. Adjust the mirror/light and focus using the low-power objective.
4. Observe the shape, size, and arrangement of the organisms.
5. Repeat the observation using a higher magnification (if required).
6. Compare both slides and note similarities and differences.

Observation Table:

Feature	Bacteria	Cyanobacteria
Cell Type	Unicellular	Unicellular (sometimes colonial)
Cell Structure	Prokaryotic	Prokaryotic
Colour	Colourless	Green/blue-green
Mode of Nutrition	Heterotrophic/Autotrophic (photosynthetic)	Autotrophic
Habitat	Everywhere (soil, water, air)	Mostly aquatic

Conclusion:

• Bacteria and cyanobacteria are unicellular prokaryotes grouped under Kingdom Monera.
• They lack a true nucleus and membrane-bound organelles.
• Cyanobacteria are photosynthetic, while bacteria show diverse modes of nutrition.
• Both play important roles in ecosystem functioning and nutrient cycling.

Viva Voce Questions

Question 1.
What are prokaryotic cells?

Question 2.
Why are bacteria placed under Kingdom Monera?

Question 3.
What is the main difference between bacteria and cyanobacteria?

Question 4.
Why are cyanobacteria called photosynthetic organisms?

Question 5. Name one useful and one harmful role of bacteria.

Activity 4: Preparation and Observation of Hay Infusion

Objective:

• To prepare a hay infusion and observe microorganisms.
• To study unicellular eukaryotes (Protista) under a microscope.
• To identify the presence of microscopic life in water.

Materials Required:

• Grass/straw/fodder
• Glass bottle
• Stagnant water/pond water
• Muslin cloth and thread
• Dropper
• Glass slide and coverslip
• Microscope
• Lab coat, gloves, mask (for safety)

Procedure:
Part A: Preparation of Hay Infusion
1. Collect a small sample of grass, straw, or fodder.
2. Take a glass bottle and fill one-fourth of it with the collected material.
3. Add stagnant or pond water to the bottle and mix well.
4. Cover the bottle with a muslin cloth and tie it with a thread.
5. Keep the bottle undisturbed for one week.

Part B: Observation Under Microscope
6. After a week, slightly open the bottle and take a drop of water using a dropper.
7. Place the drop on a clean glass slide and cover it with a coverslip.
8. Observe the slide under a microscope.
9. Look for moving microorganisms and compare them with known protists (e.g., Amoeba).

Precautions:

• Wear a lab coat, gloves, and a mask.
• The hay infusion may smell bad—handle carefully.
• Dispose of the solution properly after use.

Conclusion:

• Hay infusion contains microscopic organisms (protists).
• These organisms are unicellular eukaryotes found in water or moist places.
• Protists play an important role in Food chains, Oxygen production, and nutrient cycling

Viva Voce Questions

Question 1.
What is a hay infusion?

Question 2.
Why is stagnant water used in this activity?

Question 3.
Which type of organisms are observed in hay infusion?

Question 4.
Why are protists called unicellular eukaryotes?

Question 5.
Name one example of a protist observed under a microscope.

ACTIVITY 5: Study of Plant Groups (Kingdom-Plantae)

Objective:

• To study different plant groups based on their features.
• To analyse their adaptations for survival.
• To understand the advantages and challenges faced by each group.

Procedure:

1. Carefully study the features of each plant group given in the table: Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, Angiosperms
2. Identify key characteristics such as: Body structure, Presence of vascular tissues, Mode of reproduction
3. Analyse how these features help plants survive in their environment.
4. Note down: Advantages for survival, Exceptions or challenges faced
5. Record your observations in a tabular form.

Observation Table: Classes of Kingdom-Plantae

Plant Group	Salient Features	Advantages for Survival	Exceptions / Challenges
Thallophyta	Simple thallus body, no differentiation	Easy absorption, survival in water	Cannot live on land
Bryophyta	No vascular tissue, need water for reproduction	Adapted to moist land	Always require moisture
Pteridophyta	True roots, stems, leaves; vascular tissues present	Better transport system, live on land	Reproduction depends on water
Gymnosperms	Needle – like leaves, seeds in cones	Adapted to dry conditions, no water needed for reproduction	Seeds not enclosed in fruits
Angiosperms	Flowers, fruits, well- developed tissues	Efficient reproduction, seed protection	Depend on pollination agents

Conclusion:

• Plants show a gradual evolution from simple to complex forms.
• Advanced groups have better adaptations like: Vascular tissues, Seeds, Flowers
• These adaptations help plants survive in diverse environments, though each group still faces certain challenges.

Viva Voce Questions

Question 1.
What is the main difference between Thallophyta and Bryophyta?

Question 2.
Why are bryophytes called amphibians of the plant kingdom?

Question 3.
What advantage do vascular tissues provide to plants?

Question 4.
How are gymnosperms different from angiosperms?

Question 5.
Why are angiosperms considered the most advanced plants?

Start by reviewing these Exploration Class 9 Science Solutions Chapter 9 Atomic Foundations of Matter Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 9 Question Answer

Class 9 Science Ch 9 Atomic Foundations of Matter Question Answer

Atomic Foundations of Matter Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.
(i) How many electrons does A tend to give or take to become stable?
(ii) What kind of ion would it form?
(iii) How many electrons
(iii) How many electrons does B tend to give or take to become stable?
(iv) What kind of ion would it form?
(v) If A and B were to combine, what kind of bond would be formed?
(vi) What would be the formula for the compound thus formed?
Solution:
(i) Element A has 1 valence electron, so it tends to give electron to become stable.

(ii) It would form a positive ion called a cation (A⁺).

(iii) Element B has 6 valence electrons, so it tends to take electrons to complete its octet and become stable.

(iv) It would form a negative ion called an anion (B^2-).

(v) Since A is a metal (gives electrons) and B is a non-metal (takes electrons), they would form an ionic bond.

(vi) To balance the charges, two atoms of A will combine with one atom of B, so the formula will be A₂B.

Question 2.
An element X has six electrons in its outer shell and forms a diatomic molecule.
(i) Why would that be so?
(ii) What kind of bond would it form?
(iii) Draw the structure of the molecule it would form.
(iv) A certain other element T has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Solution:
(i) Element X has 6 electrons in its outermost shell, meaning it is short of 2 electrons to complete its octet (a stable noble gas configuration). To achieve stability, two atoms of X will mutually share two of their valence electrons with each other, forming a stable diatomic molecule (X₂).

(ii) Since it shares two pairs of electrons (a total of 4 electrons) between the two non-metal atoms, it forms a double covalent bond.

(iii) Formation of diatomic molecule X₂ (double covalent bond) is as follows

(iv) Element Y has 2 valence electrons and tends to lose them, while X tends to gain 2. They form an ionic compound by transfer of electrons.

Question 3.
You want to design a new ionic compound where the total positive charge is 6^+, and the total negative charge is 6^–. Which of the following combinations gives the correct number of ions?
(i) 2 Al³⁺ and 3 Cl^–
(ii) 3 Mg²⁺ and 1 \(P \mathrm{O}_4^{3-}\)
(iii) 2 F³⁺ and 3 O^2-
(iv) 3 Ca²⁺ and 2 \(S \mathrm{O}_4^{2-}\)
Solution:

• The correct combination is (iii) 2 F³⁺ and 3 O^2-. Positive charge 2 iron ions (Fe³⁺) provide a total charge of 2 × 3 = 6 +.
• Negative charge: 3 oxide ions (O^2-) provide a total charge of 3 × 2 = 6 -.
• Since both the positive and negative charges equal 6, this combination satisfies the requirement.

Question 4.
Choose the correct statement(s) and correct the false statement(s).
(i) Elements are made up of molecules and compounds are made up of atoms.
(ii) The molecule of a compound is always made up of two or more atoms of the same kind.
(iii) One molecule of nitrogen gas contains three nitrogen atoms.
(iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Solution:
(i) False
Correction: Elements are made up of atoms (or molecules of the same atom), and compounds are made up of molecules.

(ii) False
Correction: The molecule of a compound is always made up of two or more atoms of different kinds.

(iii) False
Correction: One molecule of nitrogen gas (N₂) contains two nitrogen atoms.

(iv) True
Water (H₂O) consists of two hydrogen atoms covalently bonded to one oxygen atom.

Question 5.
Write the chemical formulae for the following compounds.
(i) Aluminium nitrate
(ii) Calcium oxide
(iii) Ferric oxide
Solution:
(i) Aluminium nitrate: The formula of aluminium nitrate:

Formula: Al(NO₃)₃
The aluminium ion (Al³⁺ ) has a valency of 3, while the nitrate ion (NO₃) has a valency of 1. Thus, three nitrate ions are needed to balance the charge of one aluminium ion. Note the use of brackets around the polyatomic nitrate ion.

(ii) Calcium oxide: The formula of calcium oxide

Formula: CaO
Here, the valencies of the two elements are the same (2). Crossing them gives Ca₂O₂, which is simplified to the lowest common ratio and written simply as CaO.

(iii) Ferric oxide: The formula of ferric oxide

Formula: Fe₂O₃
Iron (Fe) can exhibit a valency of 2 or 3. The ‘ic’ suffix in the word ferric indicates that it uses the higher valency of 3. Crossing the valencies gives Fe₂O₃.

Question 6.
Write the formulae of the compounds formed from the following pairs of ions.
(i) Ca²⁺ and Br^–
(ii) Al³⁺ and \(C \mathrm{O}_3^{2-}\)
(iii) K⁺ and \(S \mathrm{O}_4^{2-}\)
(iv) \(N \mathrm{H}_4^{+}\) and Cl^–
Solution:
(i) Ca²⁺ and Br^–: The formula of calcium bromide

Formula: CaBr₂
The calcium ion (Ca²⁺) has a valency of 2, and the bromide ion (Br^–) has a valency of 1. Two bromide ions balance one calcium ion.

(ii) Al³⁺ and \(C \mathrm{O}_3^{2-}\): The formula of alminium carbonate

Formula: Al₂(CO₃)₃
The aluminium ion (Al³⁺) has a valency of 3, and the carbonate ion (\(C \mathrm{O}_3^{2-}\)) has a valency of 2. Brackets are required around the poly-atomic carbonate ion.

(iii) K⁺ and \(S \mathrm{O}_4^{2-}\): The formula of potassium sulphate

Formula: K₂SO₄
The potassium ion (K⁺) has a valency of 1, and the sulphate ion (\(S \mathrm{O}_4^{2-}\)) has a valency of 2. Two potassium ions balance one sulphate ion.

(iv) \(N \mathrm{H}_4^{+}\) and Cl^–: The formula of ammonium chloride

Formula: NH₄Cl
The ammonium ion (\(N \mathrm{H}_4^{+}\)) and the chloride ion (Cl^–) both have a valency of 1, resulting in a simple 1 : 1 ratio.

Question 7.
Which of the following, in the given figure, correctly represents the Cl- ion (Atomic number of chlorine = 17)?

Solution:
The correct representation is (ii).
The atomic number of chlorine is 17, so its neutral atom has an electronic configuration of (2, 8, 7).
To form a Cl^– ion, it gains one electron to complete its octet, making the configuration (2, 8, 8).
Diagram (ii) correctly shows 2 electrons in the first shell, 8 in the second, and 8 in the third shell.

Question 8.
Determine the formula unit mass of the following substances.
(i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser, which is essential for plant growth.
(ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents.
(iii) Sodium hydrogencarbonate(NaHCO₃), used to relieve acidity and helps in digestion.
Solution:
(i) Formula unit mass of ammonium nitrate (NH₄NO₃)
Atomic mass:
N = 14 u;
H = 1 u;
O = 16 u
Formula unit mass of NH₄NO₃ = (14 u × 2) + (1 u × 4) + (16 u × 3) = 80 u.

(ii) Formula unit mass of phosphoric acid (H₃PO₄)
Atomic mass:
H = 1 u;
P = 31 u;
O = 16 u
Formula unit mass of H₃PO₄ = (1 u × 3) + (31 u × 1) + (16 u × 4) = 98 u.

(iii) Formula unit mass of sodium hydrogencarbonate (NaHCO₃)
Atomic mass:
Na = 23 u;
H = 1 u;
C = 12 u;
0 = 16 u
Formula unit mass of NaHCO₃ = (23 u × 1) + (1u × 1) + (12u × 1) + (16u × 3) = 84 u.

Question 9.
Write the formulae for the compounds formed by the reaction of
(i) Magnesium and nitrogen
(ii) Lithium and nitrogen
(iii) Sodium and sulphur
(iv) Aluminium and oxygen
Solution:
(i) Magnesium and nitrogen

The formula of magnesium nitride

Formula: Mg₃N₂
Magnesium (Mg) has a valency of 2, and nitrogen (N) has a valency of 3. Crossing their valencies gives the formula Mg₃N₂.

(ii) Lithium and nitrogen

The formula of lithium nitride

Formula: Li₃N
Lithium (Li) has a valency of 1, and nitrogen (N) has a valency of 3. Three lithium atoms are needed to balance one nitrogen atom.

(iii) Sodium and sulphur

The formula of sodium sulphide

Formula: Na₂S
Sodium (Na) has a valency of 1, and sulphur (S) has a valency of 2. Two sodium atoms balance one sulphur atom.

(iv) Aluminium and oxygen

The formula of aluminium oxide

Formula: Al₂O₃
Aluminium (Al) has a valency of 3, and oxygen (O) has a valency of 2. Crossing their valencies gives Al₂O₃.

Question 10.
Complete the table by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO₃ is given as an example.

Solution:
The completed formulae for table are as follows.

	NO3–	SO42-	PO43-
NH4+	NH4NO3	(NH4)2SO4	(NH4)3PO4
Li+	LiNO3	Li2SO4	Li3PO4
Al3+	Al(NO3)3	Al2(SO4)3	AlPO4
Cu2+	Cu(NO3)2	CuSO4	Cu3(PO4)2

Question 11.
5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.
Solution:
According to the law of conservation of mass, the total mass of the reactants must be equal to the total mass of the products.
Total mass of reactants = Mass of sodium carbonate + Mass of acetic acid
Total mass of reactants = 5.3 g + 6.0 g = 11.3 g

Total mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium acetate
Total mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g
Since the total mass of the reactants (11.3 g) is equal to the total mass of the products (11.3 g), the law of conservation of mass is verified.

Question 12.
If a species has 11 protons, 12 neutrons, and 10 electrons, then
(i) what is its atomic number and mass number?
(ii) Is it neutral, a cation, or an anion? Explain.
(iii) Write its electronic configuration.
(iv) Name the species.
Solution:
(i) Atomic number = Number of protons = 11.
Mass number = Number of protons + Number of neutrons
= 11 + 12 = 23.

(ii) It is a cation.
This is because it has 11 positive protons but only 10 negative electrons, meaning it has lost one electron and carries a positive charge.

(iii) The electronic configuration of the species (with 10 electrons) is (2, 8).

(iv) The species is a sodium ion (Na⁺).

Question 13.
Two elements, A and B, have the following configurations
A: 2, 8, 5;
B: 2, 8, 7.
(i) Which element is more reactive?
(ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing.
(iii) Predict the formula of the compound they would form.
Solution:
(i) Element B is more reactive because it needs only 1 electron to complete its octet, whereas element A needs 3 electrons.

(ii) They will form covalent bonds. Since both are non-metals (A has 5 valence electrons and B has 7), they will achieve stability by sharing electrons rather than transferring them.

(iii) The formula of compound AB₃

Formula: AB₃
Element A (2,8, 5) has 5 valence electrons and needs 3 more to complete its octet, giving it a valency of 3. Element B (2, 8, 7) has 7 valence electrons and needs 1 more, giving it a valency of 1. By criss-crossing their valencies, we arrive at the formula AB₃.

Question 14.
Assertion (A): Copper sulphate conducts electricity in the molten state but not in the solid state.
Reason (R): Copper and sulphate ions are fixed in the lattice in the molten state, while in the solid state they can move freely.
Choose the correct option.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A
(iii) A is true, but R is false.
(iv) A is false, but R is true,
Solution:
(iii) A is true, but R is false.
The Assertion is true because ionic compounds like copper sulphate need free ions to conduct electricity, which only happens in molten or dissolved states.
However, the Reason is false because ions are fixed in the solid state and become free to move in the molten state (the reason states the opposite).

Question 15.
The species ²⁷Al, ⁸⁰Br^– and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Solution:
For ²⁷Al (Aluminium atom):

• Electrons = 13 (In a neutral atom, electrons = protons).
• Neutrons = Mass number – Protons
  = 12 – 13 = 14 neutrons.

For ⁸⁰Br^–(Bromide ion):

• Electrons = 35 + 1 = 36 electrons (Negative charge means it gained 1 electron).
• Neutrons = Mass number – Protons = 80 – 35 = 45 neutrons.

For ²⁰¹Hg²⁺ (Mercuric ion):

• Electrons = 80 – 2 = 78 electrons (2 + charge means it lost 2 electrons).
• Neutrons = Mass number – Protons = 201 – 80 = 121 neutrons.

Class 9 Science Chapter 9 Atomic Foundations of Matter Question Answer (InText)

Question 1.
A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left. Does this mean the law of conservation of mass is violated? Explain.
Solution:
No, the law is not violated. The ethanol reacts with oxygen from the air to form gaseous products like carbon dioxide and water vapour. Since the beaker is open, these gases escape into the atmosphere. If the reaction were carried out in a closed container, the total mass of the products would be exactly equal to the total mass of the reactants.

Question 2.
When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the law of conservation of mass?
Solution:
According to the Law of Conservation of Mass, the total mass of reactants is equal to the total mass of products.
Mass of hydrogen = 20 g
Mass of oxygen = 160 g
Total mass of reactants = 20g + 160 g = 180 g
Therefore, the mass of water formed = 180 g.

Question 3.
A compound consists of 40 % sulphur and 60 % oxygen by mass. In a sample of the same compound containing 20 g of sulphur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?
Solution:
According to the law of constant proportions, the elements in a chemical compound are always present in definite proportions by mass.
Mass percentage of sulphur = 40 %
Mass percentage of oxygen = 60 %
Ratio of sulphur to oxygen = 40 : 60 or 2 : 3
In a 20 g sample of sulphur, let the mass of oxygen be x.
\(\frac{\text { Mass of sulphur }}{\text { Mass of oxygen }}=\frac{2}{3}\)
\(\frac{20 g}{x}=\frac{2}{3}\)
x = \(\frac{20 \times 3}{2}\) = 30g
So, 30 g of oxygen must be present.

Question 4.
Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3 : 4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?
Solution:
The mass ratio of carbon to oxygen in carbon monoxide is 3:4.
This means 3 g of carbon combines with 4 g of oxygen.
For 9 g of carbon:
Oxygen = \(\frac{4}{3} \times 9\) = 12 g.
Therefore, 12 g of oxygen will combine with 9 g of carbon to form carbon monoxide.

Question 5.
The Law of Definite Proportions holds for compounds but not for mixtures. Give reason.
Solution:
This is because a compound is a pure substance where elements are chemically combined in a fixed, definite proportion by mass. On the other hand, a mixture is formed by simply mixing substances together physically without any chemical bond, so its components can be present in any ratio or proportion.

Question 6.
Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4 : 1 and 8 : 2, respectively. Do their results justify the law of constant proportions? Explain.
Solution:
Yes, their results justify the Law of Constant proportions.
Student X’s ratio of copper to oxygen = 4 : 1
Student Y’s ratio of copper to oxygen = 8 : 2
When we simplify student Y’s ratio \(\frac{8}{2}=\frac{4}{1}\)

Since, both students obtained the same simplified ratio (4 : 1), it proves that the elements in the copper oxide are always combined in the same fixed proportion by mass, regardless of the method of preparation.

Question 7.
Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.
Reason (R): According to Dalton’s atomic theory, atoms combine in a simple whole number ratio by mass to form compound.
Choose the correct option.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
(i) Both A and R are true, and R is the correct explanation of A.
The Assertion is true as it follows the law of constant proportions (2 : 16 simplifies to 1 : 8). The Reason is true because Dalton’s theory states that atoms combine in simple whole number ratios to form compounds, which explains why the mass ratio in water is always fixed.

Question 8.
Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).
Solution:
Each nitrogen atom has 5 valence electrons and needs 3 more to complete its octet. Therefore, two nitrogen atoms share three pairs of electrons to form a triple covalent bond.

Question 9.
The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).
Solution:
Fluorine has an atomic number of 9, so its electronic configuration is (2, 7). It has 7 valence electrons and needs 1 more to become stable. To achieve this, two fluorine atoms share one pair of electrons, forming a single covalent bond to create an F₂ molecule.

Question 10.
Show the formation of the following molecules
(i) Carbon dioxide (CO₂)
(ii) Hydrogen sulphide (H₂S)
(iii) Ammonia (NH₃)
Solution:
(i) Carbon dioxide (CO₂):
Carbon has 4 valence electrons, and oxygen has 6. To complete their octets, one carbon atom shares two pairs of electrons with each of the two oxygen atoms, forming two double covalent bonds (O = C = O).

(ii) Hydrogen sulphide (H₂S):
Sulphur has 6 valence electrons and hydrogen has 1. One sulphur atom shares one electron with each of the two hydrogen atoms to form two single covalent bonds (H—S—H).

(iii) Ammonia (NH₃):
Nitrogen has 5 valence electrons and needs 3 more. It shares one electron each with three hydrogen atoms, resulting in the formation of three single covalent bonds.

Question 11.
Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Solution:
Neon has an atomic number of 10, giving it an electronic configuration of (2, 8). Because its outermost shell is completely filled with 8 electrons, it is already stable. Since it has a complete octet, it has no tendency to lose, gain, or share electrons with other atoms.

Question 12.
What kind of ion will oxygen (O) form?
Solution:
Oxygen has 6 valence electrons and needs 2 more to complete its octet. Therefore, it gains 2 electrons to form a negatively charged ion called an anion, specifically an oxide ion (O^2-).

Question 13.
Fill in the blanks.
Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become ___________. Now, ___________ ion of magnesium and ___________ ions of chlorine combine to give magnesium chloride.
Solution:
Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become Cl^–. Now, one ion of magnesium and two ions of chlorine combine to give magnesium chloride.

Question 14.
Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.
Solution:
Formation of Cations:

1. Potassium (K): K (2, 8, 8, 1) loses 1 electron → K⁺ (2, 8, 8) + e^–

2. Calcium (Ca): Ca (2, 8, 8, 2) loses 2 electrons → Ca²⁺ (2, 8, 8) + 2e^–

Formation of Chlorides:

Potassium Chloride (KCl): One K atom transfers 1 electron to one Cl atom.

Calcium chloride (CaCl₂): One Ca atom transfers 1 electron each to two separate Cl atoms.

Question 15.
Illustrate how sodium sulphide (Na₂S) is formed.
Solution:
Sodium (Na) has 1 valence electron, and Sulphur (S) has 6 valence electrons. To achieve stability, two sodium atoms each lose one electron to form two Na⁺ ions. One sulphur atom gains these two electrons to form one S^2- ion. The electrostatic attraction between these oppositely charged ions results in the formation of sodium sulphide (Na₂S).

Question 16.
Name the following.
(i) CO₂
(ii) NO₂
(iii) SF₆
(iv) PCl₃
Solution:
(i) Carbon dioxide
(ii) Nitrogen dioxide
(iii) Sulphur hexafluoride
(iv) Phosphorus trichloride

Question 17.
Write the formula for the following.
(i) Sodium hydrogencarbonate
(ii) Sulphur dioxide
(iii) Ferric chloride
(iv) Cuprous oxide
Solution:
(i) Sodium hydrogencarbonate: The formula of sodium hydrogencarbonate

Thus, in sodium hydrogencarbonate, the positive sodium ion (Na⁺) and the negative polyatomic hydrogencarbonate ion (\(HC \mathrm{O}_3^{-}\)) have equal and opposite charges. The charges balance each other to form a neutral structure.

(ii) Sulphurdioxide: The formula of sulphur dioxide

Formula SO₂
Here, the valencies of the two elements are 4 and 2. Crossing them gives the ratio 2 : 4, which simplifies to 1 : 2. We arrive at the formula S₂O₄, but it is simply written using the lowest common ratio as SO₂.

(iii) Ferric chloride: The formula of ferric chloride

Formula: FeCl₃
Iron (Fe) can exhibit a valency of 2 or 3. The ‘ic’ suffix in the word ferric indicates that it uses the higher valency of 3.

(iv) Cuprous oxide: The formula of cuprous oxide

Formula: Cu₂O
Copper (Cu) can exhibit a valency of 1 or 2. The ‘ous’ suffix in the word cuprous indicates that it uses the lower valency of 1.

Question 18.
Write the formulae for the compounds formed from the following pairs of ions.
(i) Fe³⁺ and OH^–
(ii) K⁺ and \(C \mathrm{O}_3^{2-}\)
Solution:
(i) Fe³⁺ and OH^–
The formula of iron (III) hydroxide

Formula: Fe(OH)₃
The iron (III) ion (Fe³⁺) has a valency of 3, while the hydroxide ion (OH^–) has a valency of 1. Brackets enclose the polyatomic hydroxide ion.

(ii) K⁺ and \(C \mathrm{O}_3^{2 -}\)
The formula of potassium carbonate

Formula: K₂CO₃
The potassium ion (K⁺) has a valency of 1, and the carbonate ion (\(C \mathrm{O}_3^{2 -}\)) has a valency of 2.

Question 19.
What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?
Solution:
The compound contains an ionic bond. In the solid state, the ions are held in fixed positions by strong forces and cannot move. When dissolved in water, these ions become free to move and can carry an electric current.

Question 20.
Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its.
(i) formula
(ii) type of bond
(iii) electrical conductivity of its aqueous solution.
Solution:
Metal M has 2 valence electrons, so its valency is 2 +.
Oxygen has 6 valence electrons and needs 2 more, so its valency is 2 -.

(i) Formula: MO (Since both have a valency of 2, they combine in a 1 : 1 ratio).

(ii) Type of bond: Ionic bond (formed by the transfer of electrons from metal M to oxygen).

(iii) Electrical conductivity: The aqueous solution will conduct electricity because the compound releases free ions when dissolved in water.

Question 21.
Find the molecular mass of nitric acid (HNO₃).
Atomic mass — H = 1 u; N = 14 u; O = 16 u.
Solution:
Nitric acid (HNO₃)
Molecular mass = (1 × mass of H) + (1 × mass of N) + (3 × mass of O)
=(1 × 1 u) + (1 × 14u) + (3 × 16u)
= 1 u + 14 u + 48 u
= 63 u

Question 22.
Find the molecular mass of methane (CH₄).
Atomic mass — C = 12 u; H = 1 u.
Solution:
Molecular mass of methane (CH₄)
Atomic mass — C = 12 u; H = 1 u
Molecular mass of CH₄ = (12 u × 1) + (1 u × 4) = 16 u.

Question 23.
Find the formula unit mass of potassium chloride (KCl).
Atomic mass — K = 39 u; Cl = 35.5 u.
Solution:
Formula unit mass of potassium chloride (KCl)
Atomic mass — K = 39 u; Cl = 35.5 u
Formula unit mass of KCl = (39 u × 1) + (35.5 u × 1) = 74.5 u.

Question 24.
Find the formula unit mass of magnesium hydroxide, Mg(OH)₂.
Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.
Solution:
Formula unit mass of Mg(OH)₂
= (1 × mass of Mg) + (2 × mass of O) + (2 × mass of H)
= (1 × 24u) + (2 × 16u) + (2 × 1 u)
= 24 u + 32 u + 2 u = 58 u

Class 9 Science Chapter 9 Question Answer (Activities)

Activity 1: Let Us Investigate a Physical Change

Objective: To verify that there is no change in mass during a physical change (dissolution of salt in water).

Materials Required:

• 100 mL beaker (clean and dry)
• Digital weighing balance
• Water (about 50 mL)
• Common salt
• Spatula

Procedure:

1. Place a clean and dry 100 mL beaker on a digital weighing balance.
2. Set the balance reading to zero using the tare/reset button.
3. Pour about 50 mL of water into the beaker.
4. Add a spatula full of common salt to the water.
5. Record the mass shown on the weighing balance.
6. Stir or swirl the mixture until the salt dissolves completely.
7. Record the mass again after complete dissolution.

Observation:

• The salt dissolves completely in water, forming a solution.
• The mass of the solution remains equal to the total mass of salt and water taken initially.

Conclusion:

• There is no change in mass during the formation of a solution.
• This shows that dissolution of salt in water is a physical change.
• In physical changes, mass is conserved.

Viva Voce Questions

Question 1.
What is a physical change?

Question 2.
Why does the mass remain unchanged after dissolving salt in water?

Question 3.
What is meant by conservation of mass?

Question 4.
Is dissolving sugar in water also a physical change? Why?

Question 5.
Can a physical change be reversed? Give an example.

Activity 2: Let Us Investigate a Chemical Change

Objective: To study whether mass changes during a chemical reaction and understand the concept of conservation of mass.

Materials Required:

• Conical flask (100 mL)
• Balloon
• Digital weighing balance
• Vinegar (or lemon juice) (~20 mL)
• Baking soda (sodium hydrogen carbonate) (~2 g)

Procedure:

1. Place a clean, dry conical flask and a balloon on a digital weighing balance.
2. Set the balance reading to zero using the tare/reset button.
3. Pour about 20 mL of vinegar into the conical flask.
4. Take about 2 g of baking soda and put it inside the balloon.
5. Fix the balloon on the mouth of the conical flask carefully without spilling the baking soda.
6. Place the setup on the weighing balance and record the initial reading.
7. Now gently lift the balloon so that the baking soda falls into the flask and reacts with vinegar.
8. Observe the reaction (effervescence due to gas formation).
9. Place the setup again on the weighing balance and record the final reading.

Observation:

• A brisk effervescence is observed due to the release of carbon dioxide gas.
• The balloon inflates as gas is produced.
• The initial and final mass readings remain nearly the same when the system is closed.

Conclusion:

• The reaction between vinegar and baking soda is a chemical change.
• Carbon dioxide gas is formed during the reaction.
• When the system is closed (using a balloon), mass remains conserved.
• This verifies the law of conservation of mass, even in chemical changes.

Viva Voce Questions

Question 1.
What type of change occurs when vinegar reacts with baking soda?

Question 2.
Which gas is produced during this reaction?

Question 3.
Why does the balloon inflate during the experiment?

Question 4.
What is meant by the law of conservation of mass?

Question 5.
Why must the system be closed to observe conservation of mass?

Activity 3: Let Us Verify the Law – Group Activity

Objective: To verify the law of conservation of mass during a chemical reaction.

Materials Required:

• Two conical flasks (100 mL) labeled A and B.
• Digital weighing balance
• Sodium sulphate solution (1% m/v) (~10 mL)
• Barium chloride solution (1% m/v) (~10 mL)
• Marker/label

Procedure:

1. Take two clean and dry conical flasks and label them A and B.
2. Place both flasks on a digital weighing balance and set the reading to zero using the tare/reset button.
3. Pour about 10 mL of sodium sulphate solution into flask A.
4. Pour about 10 mL of barium chloride solution into flask B.
5. Keep both flasks on the weighing balance and record the total initial mass.
6. Transfer the solution from flask B into flask A and mix them carefully.
7. Observe the reaction taking place.
8. Place the flasks again on the weighing balance and record the final mass.
9. Compare the initial and final readings.

Observation:

• A white precipitate is formed when the two solutions are mixed.
• This precipitate is barium sulphate.
• The initial and final mass remain the same.

Chemical Reaction:

Sodium sulphate + Barium chloride → Barium sulphate + Sodium chloride

Conclusion:

• A chemical reaction occurs, forming a new substance (white precipitate).
• Despite the chemical change, mass remains conserved.
• This verifies the law of conservation of mass, which states that mass is neither created nor destroyed during a chemical reaction.

Activity 4: Let Us Experiment

Objective: To compare the properties of ionic and covalent compounds based on their solubility and electrical conductivity.

Materials Required:

• Samples of compounds: Camphor, Sodium chloride, Copper sulphate, Sugar, Naphthalene
• Beakers
• Water
• Kerosene
• Petrol
• Carbon/metal electrodes
• Battery (low voltage, e.g., 9V)
• Bulb
• Connecting wires
• Cardboard

Procedure:

(A) Solubility Test:

1. Take small amounts of each compound in separate test tubes/beakers.
2. Add water to each sample and observe whether it dissolves.
3. Repeat the same process using kerosene and petrol as solvents.
4. Record the solubility of each compound in all three solvents.

(B) Electrical Conductivity Test:

• Set up the electrical circuit using a battery, bulb, and electrodes.
• Test each compound in solid state by placing it between electrodes and observe whether the bulb glows.
• Dissolve each compound in water separately.
• Test the electrical conductivity of each solution using the same setup.
• Record whether the bulb glows or not in each case.

Observation (Summary Table):

Compound	Soluble in Water	Soluble in Kerosene/Petrol	Conductivity (Solid)	Conductivity (in Water)
Camphor	No	Yes	No	No
Sodium chloride	Yes	No	No	Yes
Copper sulphate	Yes	No	No	Yes
Sugar	Yes	No	No	No
Naphthalene	No	Yes	No	No

Conclusion:

1. Ionic compounds (e.g. sodium chloride, copper sulphate):

• Soluble in water
• Insoluble in kerosene/petrol
• Conduct electricity in aqueous solution but not in solid state

2. Covalent compounds (e.g., camphor, naphthalene, sugar)

• Usually insoluble in water (except sugar)
• Soluble in organic solvents (kerosene/petrol)
• Do not conduct electricity

3. Ionic compounds have high melting and boiling points, while covalent compounds generally have low melting and boiling points.

Viva Voce Questions:

Question 1.
What is the main difference between ionic and covalent compounds?

Question 2.
Why do ionic compounds conduct electricity in solution but not in solid state?

Question 3.
Why are covalent compounds generally insoluble in water?

Question 4.
Why does sugar not conduct electricity even when dissolved in water?

Question 5.
Which type of compound has a higher melting point, and why?

Start by reviewing these Exploration Class 9 Science Solutions Chapter 13 Earth as a System Energy Matter and Life Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 13 Question Answer

Class 9 Science Ch 13 Earth as a System Energy Matter and Life Question Answer

Earth as a System Energy Matter and Life Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
Choose the most appropriate option to describe the role of biogeochemical cycles in an ecosystem.
(i) To provide food directly to all organisms.
(ii) To recycle essential nutrients between biotic and abiotic components.
(iii) To create new elements for use by living things.
(iv) To remove pollutants and toxins from the organism.
Solution:
(ii) To recycle essential nutrients between biotic and abiotic components.

Biogeochemical cycles ensure that essential elements like carbon, nitrogen, and water are continuously reused and recycled between living organisms (biotic) and the non-living environment (abiotic), like soil, air, and water.

Question 2.
Which of the following is primarily responsible for the warming of the Earth?
(i) Solar radiation is immediately absorbed by carbon dioxide, which then releases it as heat.
(ii) The atmosphere’s tiny particles absorb incoming solar radiation, which directly heats the Earth.
(iii) The Earth’s surface absorbs solar radiation, which is then re-radiated and trapped by greenhouse gases.
(iv) The Earth’s environment is heated only by the solar radiation reflected by the clouds.
Solution:
(iii) The Earth’s surface absorbs solar radiation, which is then re-radiated and trapped by greenhouse gases.

Question 3.
Explain how climate change affects the water cycle. Illustrate with examples.
Solution:
Climate change disrupts the water cycle by increasing the Earth’s average temperature, which speeds up the movement of water between the land, ocean, and atmosphere.

Impacts and Examples:

1. Increased evaporation: Higher temperatures cause more water to evaporate from oceans and lakes. This leads to more water vapour in the atmosphere, which can cause more intense and frequent storms.
2. Changes in precipitation: Some areas may experience heavy, unseasonal rainfall and flooding, while other regions face extreme droughts because the timing and distribution of rain change.
3. Melting of ice: Global warming causes glaciers and polar ice caps to melt. This adds more liquid water to the oceans, causing sea levels to rise and altering the availability of fresh water.

These disruptions make the water cycle less predictable, affecting fresh water supplies for humans and wildlife.

Question 4.
Describe how albedo affects the Earth’s surface temperature and its climate.
Solution:
Albedo is the measure of how much solar radiation (sunlight) a surface reflects into space. It plays a critical role in regulating the Earth’s temperature.

1. High Albedo (Cooling):
Surfaces like ice and snow have high albedo because they reflect most of the sunlight. This helps keep the polar regions and the overall Earth cooler.

2. Low Albedo (Warming):
Darker surfaces like oceans, forests, and asphalt roads have low albedo. They absorb most of the sunlight and convert it into heat, which raises the surface temperature.

Climate Impact:
A change in albedo can create a significant impact on climate. For example, if global warming melts snow (high albedo), it exposes dark soil or water (low albedo). This causes the Earth to absorb even more heat, which leads to further melting and even higher temperatures.

Question 5.
How are mountain and valley breezes formed? Suppose there are two mountains, one covered with grass and another covered with barren rocks; would the temperature of the two mountain breezes be different? If so, how?
Solution:
The mountain and valley breezes are local winds caused by the differential heating and cooling of slopes and valleys.

1. Valley breeze (Daytime): During the day, the mountain slopes heat up faster than the valley floor. The air above the slopes becomes warm and rises, pulling cooler air up from the valley.
2. Mountain breeze (Nighttime): At night, the mountain slopes lose heat rapidly and become cooler than the valley. The cool, dense air flows down the slopes into the valley.

Comparison of Grass-covered versus Barren Rock Mountains
Yes, the temperature of the breezes would be different.

Mountain with barren rocks: Rocks have a low specific heat and low albedo. They absorb heat very quickly during the day and lose it very rapidly at night. Therefore, the breeze from a barren mountain will be much colder at night and hotter during the day.

Mountain with grass: Vegetation (grass) retains moisture and reflects more sunlight. Plants also undergo transpiration, which has a green cooling effect. Therefore, the breeze from a grass-covered mountain will be milder and more moderate in temperature.

Conclusion:
The breeze from the barren rock mountain would experience more extreme temperature shifts compared to the green mountains, making stable breeze from the grass-covered mountain.

Question 6.
You have witnessed weather phenomena, such as winds, storms, rainfall, etc. Which atmospheric layer is mainly responsible for such phenomena and what is the primary reason for its occurrence?
Solution:
The troposphere is the atmospheric layer primarily responsible for weather phenomena such as winds, storms, and rainfall. This is because it contains most of the air, along with water vapour and dust particles, which are needed for cloud formation and precipitation.

Additionally, the troposphere is heated from the Earth’s surface, causing the temperature to decrease with height. This temperature variation causes the warm air to rise and the cool air to sink, which drives the wind patterns and weather systems.

The presence of moisture in this atmospheric layer allows condensation and cloud formation, resulting in rainfall. Due to these characteristics, the troposphere becomes the most active layer for weather processes on Earth.

Question 7.
Explain the processes involved in the nitrogen cycle. How would life on Earth be affected if nitrogen were not cycled?
Solution:
The nitrogen cycle is the process by which nitrogen moves between the atmosphere, soil, and living organisms.

Main Processes:

1. Nitrogen fixation: Atmospheric nitrogen is converted into usable forms (like nitrates) by bacteria in the soil or by lightning.
2. Nitrification: Ammonia is converted into nitrates by soil bacteria, which plants absorb to grow.
3. Assimilation: Plants use these nitrates to make proteins, which are then consumed by animals.
4. Ammonification and denitrification: Decomposers break down dead matter into ammonia, and specific bacteria convert nitrates back into nitrogen gas, returning it to the atmosphere.

Impact if not cycled
If nitrogen were not cycled, life would eventually stop. Plants would be unable to make proteins and DNA, leading to their death. Since plants are the base of the food chain, all animals would also die from a lack of nutrition.

Question 8.
What are the impacts of deforestation on the Earth’s oxygen and carbon cycles? What are the other consequences of deforestation?
Solution:
Deforestation significantly disrupts both the carbon and oxygen cycles. Trees absorb carbon dioxide during photosynthesis and release oxygen into the atmosphere. When forests are cleared, this balance is disturbed- less carbon dioxide is absorbed, and less Oxygen is produced.

Additionally, burning of trees releases stored carbon back into the atmosphere, increasing greenhouse gas levels and contributing to global warming.

Deforestation also leads to soil erosion as tree roots no longer hold the soil together. It also reduces transpiration, which can decrease rainfall and alter local climate patterns. Furthermore, it destroys habitats, leading to loss of biodiversity.

Question 9.
Explain with a suitable diagram the path that carbon takes to go back to the atmosphere. You may start from plants using CO₂ from the atmosphere.
Solution:
Carbon moves from the atmosphere to living things and back again in a continuous cycle.
The Path of Carbon back to the Atmosphere:

1. Photosynthesis Plants take in CO₂ from the atmosphere to make food.
2. Respiration Both plants and animals break down food to get energy, releasing CO₂ back into the atmosphere as a byproduct.
3. Decomposition When plants and animals die, decomposers (like bacteria and fungi) break down their bodies, releasing the stored carbon back into the air as CO₂.
4. Combustion When wood or fossil fuels (formed from ancient plants) are burnt, the carbon is released back into the atmosphere as CO₂.

Conclusion Carbon returns to the atmosphere primarily through the biological processes of respiration and decomposition, as well as the physical process of burning fuels.

Question 10.
Why is an excess of CO₂ in the atmosphere considered undesirable even though it is required by plants?
Solution:
Although plants need CO₂ for photosynthesis, an excess of it is harmful to the Earth system for the following reasons

1. Global warming: CO₂ is a main greenhouse gas. An excess amount of it traps too much solar heat (infrared radiation) within the atmosphere, leading to a rise in the Earth’s average surface temperature.
2. Climate change: This temperature rise leads to unpredictable weather patterns, melting of polar ice caps, and rising sea levels.
3. Ocean acidification: Much of the excess CO₂ is absorbed by oceans, making the ocean water more acidic. This harms marine life, especially corals and shellfish.
4. Plant limits consumption of CO₂: Plants can only absorb a certain amount of CO₂. When the concentration exceeds what forests and oceans can process, the remaining gas continues to build up and heat the planet.

Conclusion:
While CO₂ is essential for life, its “excess” acts as a pollutant that destabilises the Earth’s climate.

Question 11.
How is heat lost from the surface of the Earth? What is its significance?
Solution:
The Earth loses heat through a process called terrestrial radiation. The Earth’s surface absorbs short wave solar radiation from the Sun during the day. At night or when cooling, the surface releases this energy back into the atmosphere in the form of long wave infrared radiation.

Significance:

1. Temperature balance This loss of heat prevents the Earth from becoming overheated, maintaining a steady average temperature suitable for life.
2. Greenhouse effect While some heat escapes into space, greenhouse gases trap a portion of this radiation. This keeps the Earth warm enough to prevent it from freezing at night.
3. Driving weather The movement of this heat from the surface to the atmosphere creates air currents and drives weather phenomena like winds and clouds.

The balance between heat gained from the Sun and heat lost from the surface is essential for maintaining the Earth’s thermal equilibrium.

Question 12.
If the Earth were a flat disc instead of a sphere, how would the patterns of solar radiation and temperature be different?
Solution:
The spherical shape of the Earth is responsible for uneven heating. If the Earth were a flat disc, the patterns would change as follows

1. Uniform solar angle: On a flat disc, the Sun’s rays would hit the entire surface at the same angle. This means every part of the Earth would receive the same amount of solar intensity.
2. No temperature zones: Currently, the equator is hot because rays hit vertically, and poles are cold because rays hit at a slant. On a flat disc, there would be no distinct tropical or polar zones; the temperature would be roughly the same everywhere.
3. No seasonal variation: The variation in day length and temperature that creates seasons is due to the Earth’s curvature and tilt. A flat disc would not experience these changes in the same way.

Conclusion:
The Earth would lose its diverse climate zones, leading to a uniform temperature across the entire surface, which would significantly change wind patterns and ocean currents.

Question 13.
Suppose there is a rise in atmospheric temperature on Earth. How would this affect the cryosphere, hydrosphere and biosphere?
Solution:
A rise in atmospheric temperature will significantly affect all of Earth’s spheres. In the cryosphere, higher temperatures will lead to the melting of glaciers and polar ice caps. This will add more water to the water bodies, thereby impacting the hydrosphere by raising the water levels. This will increase the risk of flooding in coastal regions. In the biosphere, these changes disrupt habitats, affecting both terrestrial and aquatic organisms. Many living organisms may lose their natural habitats, leading to reduced biodiversity. Additionally, altered rainfall patterns and water availability will impact agriculture and food production.

Question 14.
Explain how the Earth’s atmosphere helps in maintaining a suitable temperature for life to survive on the Earth.
Solution:
The atmosphere acts like a protective blanket that regulates the Earth’s temperature in the following ways

1. The greenhouse effect: Gases like Carbon Dioxide and Water Vapour trap the heat (infrared radiation) re-radiated by the Earth’s surface. This prevents the heat from escaping into space, keeping the planet warm enough for life to exist even at night.
2. Blocking excess radiation: During the day, the atmosphere prevents a large amount of the Sun’s high energy radiation from reaching the surface directly, protecting the Earth from becoming too hot.
3. Heat distribution: The movement of air in the atmosphere (winds) helps transfer heat from the warmer equatorial regions to the colder polar regions, creating more moderate climates.

Without the atmosphere, the Earth would experience extreme temperature shifts, becoming boiling hot during the day and freezing cold at night and thus making life impossible.

Question 15.
Describe the interrelationship between different spheres of the Earth. Illustrate with example how these spheres function in a delicate balance.
Solution:
The Earth is an integrated system consisting of four spheres: Atmosphere (air), Hydrosphere (water), Lithosphere (land), and Biosphere (living things). These spheres are not isolated but constantly interact through the exchange of matter and energy.

Interrelationship and Balance: The balance is maintained through biogeochemical cycles where nutrients move from one sphere to another. If one sphere is disturbed, it affects all the others.

Example: The Water Cycle:

1. Hydrosphere to atmosphere: Solar energy causes water to evaporate from oceans (Hydrosphere) into the air (Atmosphere).
2. Atmosphere to lithosphere: The water vapour condenses to form clouds and falls as rain (Precipitation) onto the land (Lithosphere).
3. Lithosphere to biosphere: Plants (Biosphere) absorb this water from the soil to grow.
4. Biosphere to atmosphere: Plants release water back into the air through transpiration, completing the cycle.

Significance: This delicate balance ensures that water and nutrients are recycled. For example, if the atmosphere warms up, it causes glaciers in the hydrosphere to melt, which floods the lithosphere and destroys habitats in the biosphere.

Conclusion: All spheres are interconnected, and the stability of the Earth system depends on their healthy interaction.

Class 9 Science Chapter 13 Earth as a System Energy Matter and Life Question Answer (InText)

Pause and Ponder

Question 1.
Visit the website given below and study the effect of the concentration of greenhouse gases on surface temperature: https://phet.colorado.edu/en/simulations/greenhouse effect.
Solution:
The concentration of greenhouse gases has a direct and significant effect on the Earth’s surface temperature.
The relationship between greenhouse gases and temperature is as follows:

1. Heat Trapping: Greenhouse gases in the atmosphere trap the heat (infrared radiation) that reflects off the Earth’s surface instead of letting it escape into space.

Direct Relationship: As the concentration of greenhouse gases increases, more heat is trapped within the atmosphere.

2. Temperature Rise: This trapped heat causes the Earth’s average surface temperature to rise.
Conclusion: Higher concentrations of greenhouse gases lead to an increase in surface temperature, a process commonly known as the Greenhouse Effect.

Question 2.
How does the cool mountain breeze benefit agricultural activity, particularly the crops and soil?
Solution:
Cool mountain breezes benefit agriculture in the following ways

• Temperature regulation: The cool air prevents the soil and crops from overheating during the day, maintaining an optimal temperature for growth.
• Moisture retention: Lower temperatures reduce the rate of evaporation from the soil, helping it retain moisture for longer periods.
• Frost prevention: In some regions, the movement of air prevents stagnant cold air from settling, which helps protect sensitive crops from frost damage.

Question 3.
What happens to the warm surface of water from the equator as it travels toward the poles? What impact does this movement have on the area?
Solution:
Warm surface water from the equator moves toward the poles via ocean currents. As it travels, it gradually loses heat to the atmosphere and becomes cooler and denser.

Impact: This movement acts as a global heat transfer system. It brings warmth to higher-latitude (polar) regions, making the climate in those areas milder and more habitable.

Question 4.
The CO₂ dissolved in the ocean is disturbed when the global temperature increases. What will happen to marine life?
Solution:
When global temperatures increase, the following happens to marine life

1. Ocean acidification: As the ocean absorbs more CO_2, the water becomes more acidic. This makes it difficult for marine animals like corals and mollusks to form their shells and skeletons.
2. Habitat loss: Increased acidity and temperature lead to coral bleaching. Since they provide food and shelter to many marine organisms, their destruction threatens the survival of many marine species.
3. Oxygen levels: Warmer water holds less dissolved oxygen, making it harder for fish and other aquatic organisms to breathe.

These changes disrupt the marine food chain and lead to a significant loss of biodiversity in the ocean.

Question 5.
What would happen to plants and animals on Earth if the biogeochemical cycles were disrupted and stopped? Explain by giving a few examples.
Solution:
If biogeochemical cycles (like the Water, Carbon, and Nitrogen cycles) stopped, the flow of essential nutrients would stop, making life on Earth impossible.

Impact on Plants and Animals:

• No nutrient recycling Nutrients would remain locked in dead organisms or the atmosphere and would not be returned to the soil for plants to use.
• Food chain collapse Since plants could not grow without recycled nutrients, animals would have no food source, leading to total extinction.

Examples of Disruption:

• Carbon cycle: If this cycle stopped, plants could not get the carbon dioxide needed for photosynthesis to produce food and oxygen.
• Water cycle: Without this cycle, there would be no tainfall. Fresh water sources would dry up, causing plants to wither and animals to die of thirst.
• Nitrogen cycle: Plants need nitrogen to make proteins. Without it, they cannot grow, which directly affects the survival of all animals that eat them.

Biogeochemical cycles are the “recycling systems” of Earth. Without them, the planet would run out of resources to support life.

Question 6.
Discuss how human activities increase the concentration of greenhouse gases in the atmosphere. What would you do as an individual to reduce the emission of greenhouse gases?
Solution:
Human activities like burning fossil fuels (coal, petrol, diesel) release large amounts of carbon dioxide into the atmosphere. Deforestation reduces the number of trees that absorb carbon dioxide. Industrial processes and vehicles also emit greenhouse gases like methane and nitrous oxide. These gases trap heat and increase global warming. As an individual, I would use public transport or cycle to reduce fuel use. I would save electricity and use energy-efficient appliances. Planting trees and reducing waste can also help lower greenhouse gas emissions.

Class 9 Science Chapter 13 Question Answer (Activities)

Activity 1: Exploring Earth’s Spheres

Objective:

• To identify the different spheres of the Earth (geosphere, hydrosphere, atmosphere, cryosphere, biosphere).
• To understand the interconnection between Earth’s spheres.
• To analyse how a change in one sphere affects others.

Materials Required:

• Given textbook figure (Earth’s surface image)
• Notebook and pen

Procedure:

1. Carefully observe the given figure of Earth’s surface.
2. Identify examples of each sphere:
   Geosphere (land, mountains, rocks)
   Hydrosphere (water bodies like lakes, rivers)
   Cryosphere (snow, glaciers, ice)
   Atmosphere (air, clouds)
   Biosphere (plants, animals, humans)
3. Circle or list one example of each sphere from the image.
4. Analyse how snow (cryosphere) melts and becomes part of the lake (hydrosphere).
5. Think about how changes like less snowfall affect water availability and plant growth.
6. Discuss how all spheres are interconnected and record your observations.

Observation Table:

Sphere	Example from Image	Description
Geosphere	Mountains, rocks	Solid part of Earth
Hydrosphere	Lake	Water bodies
Cryosphere	Snow on mountains	Frozen water
Atmosphere	Clouds, air	Layer of gases
Biosphere	Sheep, plants, human	Living organisms

Conclusion:

• The Earth is made up of interconnected spheres: Geosphere, Hydrosphere, Atmosphere, Cryosphere, and Biosphere.
• A change in one sphere affects others.
• For example: Less snowfall → less water in lakes → less plant growth → affects animals.
• This shows the importance of balance in natural systems.

Viva Voce Questions

Question 1.
What are the five main spheres of the Earth?

Question 2.
How is the cryosphere connected to the hydrosphere?

Question 3.
What is the role of the biosphere?

Question 4.
How does less snowfall affect living organisms?

Question 5.
Why are Earth’s spheres interdependent?

Activity 2: Reflection of Solar Radiation (Albedo)

Objective:

• To understand the concept of albedo (reflection of solar radiation).
• To compare how different surfaces reflect or absorb sunlight.
• To analyse how albedo affects temperature on Earth.

Materials Required:

• Textbook/table reference
• Notebook and pen
• (Optional) Internet/books for data collection

Procedure:

1. Read the given table on reflection of solar radiation.
2. Understand the meaning of albedo (reflectivity of a surface).
3. Observe the given values for snow, ice, and crushed rock.
4. Use reliable sources (books/websites) to complete the missing values.
5.  Record the completed table.
6. Analyse which surfaces reflect more and which absorb more heat.

Observation Table: Reflection of solar radiation by surfaces

Material	Albedo
1. Snow	0.80 – 0.90
2.  Ice	0.50 – 0.70
3. Crushed rock	0.25 – 0.30
4. Light coloured soil	0.30 – 0.40
5. Black soil	0.05 – 0.15
6. Ocean water	0.05 – 0.10

Conclusion:

• Surfaces like snow and ice have high albedo and reflect most sunlight, making them cooler.
• Dark surfaces like black soil and ocean water have low albedo and absorb more heat.
• Albedo plays an important role in Earth’s temperature and climate system.

Viva Voce Questions

Question 1.
What is albedo?

Question 2.
Which type of surfaces have high albedo?

Question 3.
Why does snow reflect more sunlight than soil?

Question 4.
How does albedo affect temperature?

Question 5.
Which surface absorbs more heat: ocean water or ice? Why?

Start by reviewing these Exploration Class 9 Science Solutions Chapter 11 Reproduction How Life Continues Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 11 Question Answer

Class 9 Science Ch 11 Reproduction How Life Continues Question Answer

Reproduction How Life Continues Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
A flower’s anthers are removed before it matures. Later, pollen from another plant of the same species is dusted onto its stigma and seeds are produced. Which process has been ensured here?
(i) Self-pollination
(ii) Cross-pollination
(iii) Fertilisation
(iv) Tissue culture
Solution:
(ii) Cross-pollination.

Question 2.
Arrange the following stages of sexual reproduction in plants in the correct order
(i) Pollen germination on stigma
(ii) Fertilisation
(iii) Pollination
(iv) Formation of zygote
Solution:
The correct order is
(iii) Pollination,
(i) Pollen germination on stigma,
(ii) Fertilisation,
(iv) Formation of zygote.

Question 3.
Assertion (A): The zygote formed after fertilisation immediately attaches to the uterus wall.
Reason (R): The uterus wall is always prepared to receive the zygote.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
(iv) A is false, but R is true.
R can be corrected as the zygote formed after fertilisation does not immediately attach to the uterus wall.

Question 4.
Why does asexual reproduction produce offspring that are genetically identical to the parent?
Solution:
Asexual reproduction produces genetically identical offspring because it involves only one parent and occurs through mitosis, with no mixing of genetic material.

Question 5.
Explain why the menstrual cycle stops during pregnancy.
Solution:
The menstrual cycle stops during pregnancy because the fertilised egg implants into the thickened uterine lining, which must be maintained to nourish the developing embryo. Since this lining cannot be shed, menstruation stops. The cycle also pauses completely because hormones produced during pregnancy prevent ovulation throughout the entire duration of pregnancy.

Question 6.
Why are flowers that bloom at night white or light in colour as compared to flowers that bloom during the day?
Solution:
Flowers that bloom at night are usually white or light in colour because they are pollinated by nocturnal insects like moths, and light colours are more visible in dim light or moonlight. They also produce a strong fragrance to attract and guide pollinators in the dark.

Question 7.
Why do vegetatively propagated plants tend to be more vulnerable to diseases than sexually reproduced plants?
Solution:
Vegetatively propagated plants are more vulnerable to diseases because they are genetically identical clones of the parent plant. Since all plants share the exact same genetic makeup, they have identical susceptibility to diseases and pests.

If one plant is affected, all others will be equally vulnerable because none of them carry any natural variation that protect them from vulnerable diseases.

Question 8.
If all flowers in a type of plant were only capable of self-pollination, how would it affect the genetic diversity over several generations? Explain.
Solution:

• If plants only self-pollinate, then genetic diversity will decrease over generations.
• This can make plants less adaptable to environmental changes and more prone to diseases.

Question 9.
A farmer wants to produce a large number of genetically identical plants quickly. Suggest suitable reproduction methods and explain why they are effective.
Solution:
The farmer should use vegetative propagation methods such as cutting, grafting and layering, or tissue culture. These methods use only one parent plant and rely on mitosis, producing genetically identical offspring quickly. Tissue culture is especially effective because it can produce thousands of disease-free plantlets from a small piece of plant tissue in a laboratory setting.

Question 10.
Suresh prepares slides with pollen grains in different sugar concentrations (0%, 2.5%, 5%, 7.5%, 10%) to study the germination of pollen.
(i) What are the different hypotheses which can be tested using this set-up?
(ii) What parameters should be kept the same in this set-up?
Solution:
(i) The hypotheses that can be tested are that pollen germination depends on sugar concentration, there is an optimum concentration for maximum germination, no germination occurs at 0% sugar, and very high concentrations may inhibit germination.

(ii) The parameters to be kept the same in this set-up are temperature, humidity, time of observation, type and source of pollen, volume of solution, light conditions and microscope magnification.

Question 11.
Look at the picture given below and find out which type of pollination might have been followed in these flowers.

Solution:
Tomato:
Tomato uses self-pollination because the stamens surround and completely cover the stigma, making it easy for pollen to fall directly onto it without any external agent being needed.

Wheat:
Wheat uses self-pollination because pollination occurs within the same flower, often before the flower fully opens, leaving little chance for external agents.

Papaya:
Papaya uses cross-pollination because male and female flowers grow on completely different trees, making self-pollination entirely impossible and requiring pollen to be transferred between trees.

Question 12.
In the lower Himalayan region of northern India, apples are an important cash crop that contribute significantly to farmer’s livelihoods. The fruit yield in apple cultivation is declining continuously, associated with climate change and a significant decline in the population of natural pollinators. A researcher-farmer group set up two experimental apple orchards at two distinct locations: Places A and B. In apple orchards at Place A, they allowed natural pollinators to pollinate the flowers of the apple. In apple orchards at Place B, they applied mixed farming techniques of beekeeping. Along with honey, the farmer yielded apples. The yield of apples is depicted in figure in terms of fruit setting (number of fruits / total number corresponding of fruit-bearing branches) and fruit drop (premature falling of developing fruits) in the two types of experimental places of apple orchards.

(i) What are the hypotheses the researcher-farmer group has thought of for this investigation?
(ii) What are the different parameters in the experiment?
(iii) Compare and analyse the data of two experimental orchards Places A and B in terms of high yields of apple fruits.
(iv) Based on your analysis, what do you infer from the data?
Solution:
(i) Hypotheses for the investigation:

• The presence of managed bee colonies increases pollination efficiency.
• Better pollination leads to higher fruit setting and lower fruit drop in apple orchards.
• Decline in natural pollinators reduces apple yield.

(ii) Parameters in the experiment:

• Independent variable: Type of pollination (natural pollination vs pollination with bee colonies).
• Dependent variables: Fruit setting rate (number of fruits formed per branch), Fruit drop rate (premature falling of fruits).
• Controlled variables: Type of apple plants, environmental conditions, soil, water, and farming practices (except beekeeping).

(iii) Comparison and analysis:
Place A (natural pollination):

• Lower fruit setting
• Higher fruit drop
• Overall lower yield

Place B (with bee colonies):

• Higher fruit setting
• Lower fruit drop
• Overall higher yield
• Place B shows better apple production than Place A.

(iv) It can be inferred from the given data that the use of bee colonies improves pollination, resulting in higher fruit production and reduced fruit loss. Therefore, beekeeping is an effective method to increase apple yield, especially where natural pollinators are declining.

Question 13.
A student claims, “In humans, ovulation always happens on day 14 of the menstrual cycle.” Critically examine this claim and state whether the claim is correct or not. Give at least two reasons for your answer.
Solution:
The claim is not entirely correct because ovulation on day 14 is only an approximation based on a typical 28-day menstrual cycle.

Firstly, the length of the menstrual cycle varies from about 21 to 35 days in different individuals, so ovulation does not always occur on day 14. Secondly, factors such as stress, illness and hormonal imbalances can alter the timing of ovulation, making it variable rather than fixed.

Class 9 Science Chapter 11 Reproduction How Life Continues Question Answer (InText)

Think it Over

Question 1.
When does a farmer prefer asexual or sexual methods of reproduction for crop production?
Solution:
A farmer prefers asexual reproduction when he wants genetically identical plants, faster results, or to propagate plants without seeds, such as bananas and seedless grapes.
He prefers sexual reproduction when developing new varieties with better characteristics through hybridisation or when growing seed crops like wheat and rice.

Question 2.
Why do most complex animals and flowering plants use sexual reproduction, while simple organisms like yeast and hydra mainly reproduce asexually?
Solution:
Complex animals and flowering plants use sexual reproduction because it creates genetic variation through meiosis and the mixing of chromosomes from two parents. This variation helps species adapt to changing environments and contributes to evolution. Simple organisms like yeast and hydra reproduce asexually because they live in stable environments where cloning is sufficient for survival.

Pause and Ponder

Question 1.
In a china-rose (Hibiscus or gudhal) plant, a pollen tube grows and continues through the style after pollen lands on the stigma. Which process is about to happen next?
Solution:
The process about to happen next is fertilisation. The pollen tube carries the male gamete down through the style into the ovary. The male gamete then fuses with the egg cell inside the ovule. This fusion of gametes is called fertilisation, and the resulting fertilised egg is called a zygote

Question 2.
Look at the pictures of calotropis (madar) seeds and dandelion seeds. Can you guess what kind of seed dispersal these seeds are adapted for?

Solution:
Calotropis and dandelion seeds are adapted for wind dispersal with lightweight structures that help them float in air. Calotropis has silky fibres and dandelion has a pappus, enabling seeds to travel long distances before germination.

Question 3.
A farmer plants two varieties of maize side by side, but notices that seeds form only when pollen from one variety reaches the stigma of the other. What type of pollination is this?
Solution:
This is an example of cross-pollination, where pollen from one plant reaches the stigma of another plant of the same species.

Question 4.
Why do animals with external fertilisation generally produce more eggs than animals with internal fertilisation?
Solution:
Animals with external fertilisation produce more eggs because fertilisation takes place outside the body in water where many eggs and sperm are destroyed by water currents, temperature changes or predators.

Question 5.
In animals, which fertilisation method the gametes are more protected?
Solution:
Internal fertilisation gives more protection to gametes as fertilisation occurs inside the female body, safeguarding them from external hazards.

The fertilised egg develops in a protected environment, leading to higher survival rates and fewer eggs produced.

Question 6.
Ravi suddenly notices that he is growing taller rapidly, his shoulders are broadening, and his voice cracks. What stage of life is he entering?
Solution:
The stage of life Ravi is entering is puberty. It is a stage of adolescence marked by rapid physical and hormonal changes, leading to maturation of reproductive organs and sperm production.

Question 7.
Rina’s period occurs every 28 days. Her last period was on the 5th of March. On which day is she most likely to get her next period?
Solution:
Rina’s next period is most likely to begin on 2nd April. Since her cycle repeats every 28 days and her last period began on 5th March, adding 28 days gives 2nd April.

Question 8.
A human zygote has just formed. How many chromosomes does it have?
Solution:
A human zygote has 46 chromosomes (23 pairs).

Question 9.
What protective devices can be used during sexual activity to reduce the spread of STIs?
Solution:
Condoms are the most effective protective devices used during sexual activity to reduce the spread of STIs.

They act as a physical barrier preventing contact between body fluids, thereby blocking infections like gonorrhoea, herpes, syphilis and HIV, while also preventing unwanted pregnancy.

Question 10.
If a couple uses oral contraceptive pills but not condoms, which risks remain and why?
Solution:
The risk of STIs remains because oral contraceptive pills only alter hormone levels to prevent ovulation and do not provide a physical barrier. Thus, exchange of body fluids continues, allowing transmission of infections like HIV, herpes, and syphilis.

Question 11.
In many animals, the young ones can walk or find food soon after birth, but human babies are completely dependent for a long time. What might be some advantages and disadvantages of this for humans as a species?
Solution:
Advantage: This long dependency allows human babies to develop their brain, learn language, skills, and complex social behaviour, benefiting humans as a species.
Disadvantage: Human babies remain dependent for a long time, requiring constant care, which increases the risk to survival if proper care is not available.

Class 9 Science Chapter 11 Question Answer (Activities)

Activity 1: Let Us Explore (Vegetative Propagation Methods)

Objective: To observe and understand different methods of asexual reproduction in plants—cutting, grafting, and layering—and their role in plant propagation.

Materials Required:

• Healthy plant stems/twigs
• Garden soil mixed with compost
• Water
• Knife/blade (for cutting and grafting)
• Cotton cloth/polythene strip (for grafting)
• Potted plants or field access (garden/farm)

Procedure:

(A) Cutting:

1. Observe how gardeners cut overgrown branches during the growing season.
2. Prepare cuttings from a healthy plant stem.
3. Note:
   Length of the cutting
   Number of nodes and internodes
4. Remove leaves from the lower half of the cutting.
5. Insert the cutting into soil at an angle (about 45°-60°).
6. Water regularly and observe growth of new shoots.

(B) Grafting:

1. Take a healthy rooted plant (Plant A – stock).
2. Take a healthy stem cutting from another plant (Plant B – scion).
3. Make a slit on the stem of Plant A.
4. Insert the cutting (scion) of Plant B into the slit.
5. Tie and cover the joint with cloth/polythene to protect it.
6. Water regularly and observe growth after healing.

(C) Layering:

1. Select a flexible twig of a plant (e.g., lemon plant).
2. Bend the twig and bury its middle portion in soil.
3. Water regularly.
4. After 10-15 days, observe root development from the buried part.
5. Cut the twig from the parent plant once roots develop.

Observation:

• In cutting, new plants grow from stem pieces.
• In grafting, parts of two plants grow together as one plant.
• In layering, roots develop from a branch while still attached to the parent plant.

Conclusion:

• Cutting, grafting, and layering are methods of asexual reproduction (vegetative propagation).
• These methods help in producing new plants identical to the parent plant.
• They are widely used in agriculture and horticulture for rapid and controlled plant growth.

Viva Voce Questions

Question 1.
What is vegetative propagation?

Question 2.
Which method involves two different plants?

Question 3.
Why are leaves removed from the lower part of a cutting?

Question 4.
How is layering different from cutting?

Question 5.
Why are these methods useful for farmers and gardeners?

Activity 2: Let Us Explore (Budding in Yeast)

Objective: To observe the process of budding in yeast and understand asexual reproduction in microorganisms.

Materials Required:

• Sugar solution (1 g sugar in 10 mL water)
• Test tube
• Yeast granules
• Cotton plug
• Glass slide and coverslip
• Dropper
• Compound microscope

Procedure:

1. Take about 20 mL of sugar solution in a test tube.
2. Add a pinch of yeast granules to the solution.
3. Close the mouth of the test tube with a cotton plug.
4. Keep the test tube in a warm place for 1-2 hours to activate the yeast.
5. After some time, take a drop of the yeast mixture and place it on a glass slide.
6. Cover it with a coverslip carefully.
7. Observe the slide under a compound microscope at different magnifications.
8. Draw a diagram of what you observe.

Observation:

• Small, round outgrowths (buds) are seen on the yeast cells.
• These buds gradually grow in size.
• Some buds detach and form new yeast cells.

Conclusion:

• Yeast reproduces by a method called budding.
• In this process, a small outgrowth develops on the parent cell.
• The bud grows and eventually separates to form a new individual.
• This is a type of asexual reproduction.

Viva Voce Questions

Question 1.
What is budding?

Question 2.
Why is yeast kept in a warm place during the experiment?

Question 3.
What is the role of sugar solution in this activity?

Question 4.
What do the small outgrowths on yeast cells represent?

Question 5.
Name another organism that reproduces by budding.

Activity 3: Let Us Explore (Variation and Combination of Characters)

Objective: To understand how different combinations of characters are formed during reproduction and how variation arises.

Materials Required:

• Beads of different colours (in pairs)
• Green (light and dark)
• Blue (light and dark)
• Red (light and dark)
• Container or box for random selection
• Notebook for recording combinations

Procedure:

1. Take three pairs of coloured beads:
   Pair 1 (green): light green and dark green
   Pair 2 (blue): light blue and dark blue
   Pair 3 (red): light red and dark red
2. Each pair represents contrasting characters (traits) on chromosomes.
3. Randomly pick one bead from each pair.
4. Record the combination obtained (e.g., light green, light blue, dark red).
5. Repeat the activity multiple times and note different combinations.
6. Count the total number of possible combinations

Observation:

• Each selection gives a different combination of traits.
• With three pairs, a total of 8 combinations (23) are possible.
• Combinations are formed randomly.

Conclusion:

• Different combinations of traits arise due to random segregation of chromosomes during gamete formation.
• This leads to genetic variation among individuals.
• Variation ensures that no two individuals (except identical twins) are exactly alike.

Viva Voce Questions

Question 1.
What is meant by variation?

Question 2.
Why do different combinations of traits occur?

Question 3.
How many combinations are possible with three pairs of characters?

Question 4.
What is the role of meiosis in variation?

Question 5.
Why are siblings not identical to each other?

Question 6.
What does the term ‘eukaryotic’ mean, and which types of cells are classified under it?

Activity 4: Let Us Investigate (Pollination in Pea Plant)

Objective: To study the role of stamens and pollination in the formation of fruits in a pea plant.

Materials Required:

• Pea plant (sweet pea/garden pea)
• Muslin cloth bags
• Thread or string
• Forceps/Scissors (to remove stamens)
• Notebook for recording observations

Procedure:

1. Identify a pea plant in a garden or nearby field.

2. Select:

• • Two flower buds (juvenile stage)
  • Three freshly bloomed flowers on the same plant

3. Carefully remove the stamens from

• • One flower bud
  • One fully bloomed flower

4. Cover the following flowers with muslin cloth bags

• • A normal flower bud
  • A flower bud with removed stamens
  • A flower with removed stamens
  • A normal flower

5. Leave one flower uncovered (without muslin cloth).
6. Water the plant regularly and observe the development of fruits.
7. After a few days, remove the muslin cloth and observe the results.
8. Record your observations in the table.

Observation Table:

Treatment	Fruit Formation (Yes/No)
Flower bud (wrapped with muslin cloth bag)	Yes
Flower bud with removed stamens (wrapped)	No
Flower with removed stamens (wrapped)	Yes
Flower (wrapped with muslin cloth bag)	Yes
Flower (without muslin cloth bag)	Yes

Observation:

• Fruits are formed in most flowers except the one where stamens were removed at the bud stage.
• Covering flowers prevents external pollination.
• Removal of stamens prevents self-pollination in buds.

Conclusion:

• Pollination is necessary for fruit formation.
• Stamens produce pollen required for fertilisation.
• If stamens are removed before maturity, pollination does not occur, and fruits are not formed.
• This activity demonstrates the importance of pollen transfer in the reproduction of plants.

Viva Voce Questions

Question 1.
What is pollination?

Question 2.
What is the function of stamens in a flower?

Question 3.
Why are flowers covered with muslin cloth in this activity?

Question 4.
Why does fruit not form when stamens are removed from the bud?

Question 5.
What is the difference between self-pollination and cross-pollination?

Start by reviewing these Exploration Class 9 Science Solutions Chapter 10 Sound Waves Characteristics and Applications Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 10 Question Answer

Class 9 Science Ch 10 Sound Waves Characteristics and Applications Question Answer

Sound Waves Characteristics and Applications Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
Which observation best supports the idea that sound is a mechanical wave?
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
Solution:
(ii) A mechanical wave is defined as a wave that requires a material medium (like air, water, or solids) to travel because it relies on the vibration of particles to transfer energy. While reflection, frequency, and energy are properties of many types of waves (including light, which is electromagnetic), the strict requirement for a medium is the defining characteristic of a mechanical wave.

Question 2.
For a sound wave propagating in a medium, increasing its frequency will increase its
(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
Solution:
(iii) When we increase the frequency, we are directly increasing the number of compressions passing a point per second. Wavelength decreases as frequency increases (inverse relationship). Speed remains constant as it depends on the medium, not the frequency. Time period decreases as frequency increases (T = 1/f).

Question 3.
If 20 compressions pass a point in 4 s, the frequency is
(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
Solution:
(ii) Frequency is defined as the number of oscillations (or compressions) that pass a given point per second.
Number of compressions = 20
Time taken = 4 s
Frequency (f) = \(\frac{\text { Total compressions }}{\text { Total time }}\)
f = \(\frac{20}{4}\) = 5 Hz

Question 4.
In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.
Solution:
No, it will not produce an echo, but only a reverberation. For an echo to be heard clearly, the time gap between the original sound and the reflected sound must be at least 0.1 s. Here, the reflected sound returns in 0.05 s, which is less than 0.1 s. So, the sounds overlap and are heard as reverberation instead of a separate echo.

Question 5.
Graphs representing two sound waves are given in the figure. If the scales on the X and F-axes of the two graphs are the same, which of the two sound waves has
(i) greater wavelength, and
(ii) smaller amplitude?

Solution:
(i) In graph (a), the peaks are spread further apart along the X-axis (distance) compared to graph (b), where the waves are tightly packed. Thus, sound wave (a) has greater wavelength.

(ii) Visually, the wave in graph (a) is shorter in height compared to the wave in graph (b), which has taller peaks and deeper valleys. Thus, graph (a) has smaller amplitude too.

Question 6.
The sound waves emitted by three sources A, B and C are represented in the figure below. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Solution:
On a density-distance graph, higher frequency means the waves are more crowded (closely packed), while lower frequency means the waves are more spread out.

Source-A (Maximum Frequency):
This corresponds to the green curve. It has the most cycles (peaks and valleys) within the same distance, indicating the highest frequency.

Source-B (Intermediate Frequency):
This corresponds to the red curve. It has fewer cycles than the green wave but more than the blue wave.

Source-C (Minimum Frequency):
This corresponds to the blue curve. It is the most spread out, with only two full cycles shown, indicating the lowest frequency.

Question 7.
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Solution:
To represent this wave on a density-distance graph

• Amplitude (3 units): The maximum height of the wave (peak) and the maximum depth (trough) from the centre line should be 3 units on the vertical Y-axis.
• Wavelength (4 cm): The horizontal distance for one complete cycle (one peak and one trough) should be exactly 4 cm on the horizontal X-axis.

Question 8.
In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
Solution:
There are two major scientific errors in this scene
(i) Sound cannot travel in space:
Sound is a mechanical wave that requires a material medium (like air or water) to propagate. Since space is a vacuum, there are no particles to vibrate and carry sound waves. Therefore, an explosion in space would be completely silent.

(ii) Difference in speeds:
Even if a medium were present, light (c ≈ 3 × 10⁸ m/s ) travels much faster than sound (v ≈ 344 m/s). In any explosion at a distance, you would always see the flash of light first and hear the sound much later. Showing them at the same time is physically inaccurate.

Question 9.
A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 ms^-1 ? find its time period.
Solution:
Given,
Wavelength (λ) = 3.44 m
Speed of wave (v) = 344 m/s
We know that,
f = \(\frac{υ}{\lambda}\)
f = \(\frac{344}{3.44}\) = 100 Hz
Also, T = \(\frac{1}{f}\)
= \(\frac{1}{100}\) = 0.01 s

Question 10.
A ship searching for a sunken ship sent a SONAR signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 ms^-1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Solution:
When using SONAR, the sound travels down to the object and then reflects back up. Therefore, the total distance travelled is twice the depth (2d) of the ocean.

Given, speed of sound (v) = 1525 m/s
Total time for echo(t) = 5 s
Total distance = Speed × Time
2d = v × t
2d = 1525 × 5 =7625
d = \(\frac{7625}{2}\) = 3812.5 m

Question 11.
A vehicle is fitted with an ultrasonic distance sensor as part of a parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by the ultrasonic wave to travel to the obstacle and come back? Assume the speed of the ultrasonic wave in air to be 345 ms^-1.
Solution:
In an echolocation system, the sound travels to the obstacle and reflects back. Therefore, the total distance (D) travelled is twice the distance to the obstacle (d).

Distance to the obstacle (d) = 1.2 m
Speed of sound (v) = 345 m/s
Total distance, d = 2d
= 2 × 1.2 = 2.4 m
Time taken, t = \(\frac{\text { Total distance }}{\text { Speed }}\)
= \(\frac{D}{v}\)
= \(\frac{2.4}{345}\) ≈ 0.00696 s
Thus, the time taken by the wave is 0.00696 s.

Question 12.
The speed of sound in air is about 331 m/s at 0°C and nearly 344 m/s at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
Solution:
Given, distance = 1720 m
Speed of sound at 22 °C = 344 m/s
Speed of sound at 0 °C = 331 m/s
Time taken at 22 °C,
t₁ = \(\frac{1720}{344}\) = 5 s
Time taken at 0 °C
t₂ = \(\frac{1720}{331}\) ≈ 5.196 s
Extra time = t₂ – t₁
= 5.196 – 5 = 0.196 s

Question 13.
The variation of density of medium for a sound wave propagating with a speed of 340 ms^-1 is shown in figure. Calculate the wavelength and frequency of the sound wave.

Solution:
In the diagram, the distance of 8 cm covers two full cycles (from the center of one compression to the center of the third compression). Therefore, one wavelength (λ) is half of that distance.
Also, speed of sound (v) = 340 m/s 8cm
λ = \(\frac{8 cm}{2}\)
= 4 cm = 0.04 cm
Using wave formula,
f = \(\frac{υ}{\lambda}\)
= \(\frac{340}{0.04}\)
= 8500 Hz

Question 14.
The graphical representation of two sound waves A and B propagating at the same speed of 345 m/s is shown in figure. What is the wavelength of each of them? Also, calculate their frequencies.

Solution:
Wavelength of wave A = 2.5 cm = 0.025 m
Wavelength of wave B = 5 cm = 0.05 m
From the formula
v = \(\frac{v}{\lambda} \)
Wave A, v_A = \(\frac{345 \mathrm{~m} / \mathrm{s}}{0.025 \mathrm{~m}} \) = 13,800 Hz
Wave B, v_B = \(\frac{345 \mathrm{~m} / \mathrm{s}}{0.05 \mathrm{~m}}\) = 6,900 Hz

Question 15.
Two identical sound sources are placed at A and B—one in air and one submerged in water figure. Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?

Solution:
Let the distance between the sound sources and cliff be d.
We know that,
v = \(\frac{d}{t}\),
so v ∝ \(\frac{1}{t}\)
Let t_A and t_B be the time taken by source at A and B to travel and return.
So, according to question,
t_A = 4.5 × t_B
Also, the ratio of the speed of sound in air (v_A) to the speed of sound in water (v_B) is the inverse of their time ratio.
\(\frac{v_A}{v_B}=\frac{t_B}{t_A}\)
Substituting the value of t_a in the equation,
\(\frac{v_A}{v_B}=\frac{t_B}{4.5 \times t_B}=\frac{1}{4.5}\)
\(\frac{1 \times 2}{4.5 \times 2}=\frac{2}{9}\)
Thus, the ratio of speed of sound in air to water is 2/9.

Class 9 Science Chapter 2 Cell The Building Block of Life Question Answer (InText)

Question 1.
Explore various ways of producing sound.
Solution:
Sound can be produced in different ways, such as:

• By plucking a string (guitar, sitar)
• By blowing air into an instrument (flute, trumpet)
• By striking or hitting objects (drum, bell)
• By vibrating vocal cords (human voice)
• By shaking or rubbing objects (rattles, violin bow)

Question 2.
Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Solution:
Musical instruments are generally categorised by how they create sound.

Instrument Type	Examples	Vibrating Part (Sound Source)
Stringed	Veena,Guitar,Violin	Stretched strings
Wind	Flute, Trumpet, Shehnai	Air column inside the instrument
Percussion	Tabla, Dholak, Drum	Stretched membrane (skin)
Solid/Idiophones	Manjira (Cymbals), Ghatam	The entire body of the instrument
Reed	Harmonium, Saxophone	A thin piece of wood or metal (reed)

Question 3.
Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
Reason (R): Sound requires a medium to travel.
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
(ii) Sound is a mechanical wave that requires a medium to propagate, meaning it cannot travel through the vacuum created by removing the air. Since, the lack of a medium is the direct cause for the bell being inaudible, the reason perfectly explains the assertion.

Question 4.
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
(iii) While compressions and rarefactions represent the energy of the wave traveling forward, the individual particles of the medium only vibrate back and forth about a fixed mean position. Because the particles do not actually travel along with the wave to the destination, the reason provided is scientifically incorrect.

Question 5.
When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
(i) Air particles near the tuning fork.
(ii) Energy carried by sound waves.
(iii) The tuning fork material.
(iv) A continuous stream of compressed air.
Solution:
(ii) Sound waves transfer energy through a medium by causing particles to vibrate in place rather than travelling from the source to the listener.
Therefore, while the air particles stay in their general area, it is the disturbance or energy they pass along that actually reaches your ear.

Question 6.
The variation of density of the medium for two sound waves is shown in figures (a) and (b) given below. Label compression and rarefaction by C and R on it. In the graph given in figures (c) and (d), label the axes and draw the curves corresponding to figures (a) and (b).

Solution:
On figures (a) and (b), the dark, crowded regions are C (Compression) and the light, spread-out regions as R (Rarefaction). For the graphs (c) and (d), the Y-axis is Density/Pressure and the X-axis as Distance/Time, required graph is drawn below.

Question 7.
Conduct activity with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?
Solution:
Yes, the thin rubber band vibrates faster than the thick one. It has a higher frequency (more vibrations per second) and a smaller time period (takes less time to complete one vibration).

Question 8.
If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?
Solution:
Frequency is 20 oscillations per second.
In one minute (60 seconds):
20 oscillations/second × 60 seconds = 1200 oscillations/minute

Question 9.
For the sound wave represented by the graph shown in the figure, what is half of its wavelength?

Solution:
According to the graph, one full wave cycle (one wavelength) spans a distance of 3.0 cm.
Therefore, half of the wavelength is 3.0 cm ÷ 2 = 1.5 cm.

Question 10.
Two friends are standing along a steel fence at a distance of 340 m from each other. Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)

Solution:
Given,
Distance d = 340 m
Speed in air, v_air = 344 m/s
Speed in steel, v_steel = 5960 m/s
Time taken by sound to travel in air,
t_air = \(\frac{d}{v_{\text {air }}}\)
= \(\frac{340}{344}\) ≈ 0.988 s
Time taken by sound to travel in steel,
t_steel = \(\frac{d}{v_{\text {air }}}\)

The sound travels much faster through the steel fence than through the air, arriving at Gunjan’s ear almost a full second earlier via the metal.
Time difference = t_air – t_steel
= 0.988 – 0.057 = 0.931 s.

Because the resulting time gap of 0.93 seconds is significantly higher than the 0.1 second human hearing threshold, Gunjan will hear two distinct knocks rather than one combined sound.

Question 11.
An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 ms^-1.
Solution:
Given,
Speed of sound, v = 343 ms^-1
Total time for echo, t = 0.2 s
For an echo, the sound covers the distance to the reflector and back.
Total distance travelled,
2d = v × t
2d = 343 × 0.2 = 68.6 m
Minimum distance to surface (d) = \(\frac{68.6}{2}\) = 34.3 m
The reflecting surface must be placed at a minimum distance of 34.3 m.

Question 12.
Sound travels much farther in water than light, and thus is used for various underwater applications. A sonar signal sent to find the depth of the ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 ms^-1 ?
Solution:
Given,
Speed of sound in seawater, v = 1500 ms^-1
Time taken for the signal to return, t = 4 s
For SONAR, the signal travels to the seabed and back, covering the depth twice.
2d = v × t
Total distance travelled,
2d = 1500 × 4 = 6000 m
Depth of the ocean, d = \(\frac{6000}{2}\) = 3000 m
The depth of the ocean is 3000 m (or 3 km).

Class 9 Science Chapter 2 Question Answer (Activities)

Activity 1: Sound Produced by Vibrations

Objective:

• To understand that sound is produced by vibrations.
• To observe the relationship between vibration and sound.
• To study how tension affects sound.

Materials Required:

• Cardboard box (one side open)
• Rubber band
• Fingers (for plucking)
• Procedure

Procedure:

1. Take a cardboard box with one side open.
2. Stretch a rubber band across the open side of the box.
3. Hold the box firmly and pluck the rubber band with your finger.
4. Observe whether any sound is produced.
5. Pluck the rubber band again and observe its motion.
6. Wait until the rubber band stops vibrating and check if the sound continues.
7. Now change the tension in the rubber band by tightening or loosening it and pluck again.
8. Remove the rubber band from the box, stretch it between two fingers, and pluck it near your ear.

Observation:

• When the rubber band is plucked, it vibrates and produces sound.
• When the vibration stops, the sound also stops.
• Increasing tension produces a different (usually higher pitch) sound.
• The sound is louder when the rubber band is stretched over the box due to resonance.

Conclusion:

• Sound is produced due to vibrations of objects.
• Vibrations are necessary for sound production.
• The tension of the vibrating object affects the sound produced.
• The object that produces sound is called the source of sound.

Viva Voce Questions

Question 1.
What is the main cause of sound production?

Question 2.
What happens to sound when vibrations stop?

Question 3.
How does tension affect the sound produced?

Question 4.
Why is the sound louder when the rubber band is on the box?

Question 5.
What is meant by vibration?

Activity 2: Sound Produced by a Vibrating Tuning Fork

Objective:

• To demonstrate that sound is produced by vibrating objects.
• To observe vibrations using a tuning fork.
• To relate vibrations to wave formation in water.

Materials Required:

• Tuning fork
• Soft rubber pad
• Bowl of water

Procedure:

1. Take a tuning fork and hold it by its stem.
2. Gently strike one of its prongs against a soft rubber pad.
3. Bring the vibrating tuning fork close to your ear and listen to the sound.
4. Now, gently touch one prong of the vibrating tuning fork to the surface of water in a bowl.
5. Observe the formation of waves on the water surface.
6. Repeat the activity by holding the tuning fork in different positions and observe the sound each time.

Observation:

• A sound is heard when the tuning fork is struck.
• The prongs of the tuning fork are vibrating rapidly.
• When the vibrating prong touches water, ripples (waves) are formed.
• This confirms that the tuning fork is vibrating.

Conclusion:

• Sound is produced due to vibrations of objects.
• Vibrations can be observed through their effects (like water waves).
• Thus, a tuning fork acts as a source of sound because of its vibrations.

Viva Voce Questions

Question 1.
What happens when a tuning fork is struck?

Question 2.
Why should the tuning fork not be struck against a hard surface?

Question 3.
What do the waves in water indicate?

Question 4.
Why do we hear sound when the tuning fork vibrates?

Question 5.
What is the source of sound in this activity?

Activity 3: Propagation of Sound through Liquids

Objective:

• To understand that sound can travel through liquids.
• To observe the propagation of sound through different media.
• To learn that a medium is necessary for sound transmission.

Materials Required:

• Large tub or bucket filled with water
• Two metal spoons

Procedure:

1. Take a bucket filled with water.
2. Hold two metal spoons and tap them against each other in air.
3. Listen carefully to the sound produced.
4. Now, immerse both spoons in water without touching the sides or bottom of the bucket.
5. Tap the spoons together again underwater.
6. Listen carefully and compare the sound heard in both cases.

Observation:

• Sound is clearly heard when spoons are tapped in air.
• Sound is also heard when spoons are tapped underwater.
• This shows that sound reaches the ear through both water and air.

Conclusion:

• Sound can travel through liquids as well as gases.
• A material medium is required for sound to travel.
• Sound cannot travel in a vacuum due to the absence of a medium.

Viva Voce Questions

Question 1.
What is a medium in sound propagation?

Question 2.
Can sound travel through liquids? Give an example.

Question 3.
Why is sound not heard in a vacuum?

Question 4.
Through which mediums can sound travel?

Question 5.
What did you observe when spoons were tapped under water?

Activity 4: Let Us Investigate Propagation of Disturbance in a Slinky (Longitudinal Waves)

Objective:

• To understand how disturbances travel through a medium.
• To observe compressions and rarefactions in a longitudinal wave.
• To distinguish between the movement of particles and the movement of the disturbance.

Materials Required:

• Slinky (spring)
• Marker
• Flat surface (table or floor)
• One helper (friend)

Procedure:

1. Take a slinky and mark one of its turns using a marker.
2. Stretch the slinky horizontally on a table or floor.
3. Ask a friend to hold one end of the slinky while you hold the other end.
4. Give a sharp push and pull to your end of the slinky.
5. Observe the disturbance created and its movement towards the other end.
6. Repeat the push and pull multiple times in quick succession.
7. Carefully observe:

• • Movement of the marked point
  • Formation of closely spaced and widely spaced regions

Observation Table:

• Regions where coils are closer together are called compressions.
• Regions where coils are spread apart are called rarefactions.
• The disturbance travels along the slinky from one end to the other.
• The marked point does not travel with the disturbance; it only moves back and forth about its mean position.

Conclusion:

• The disturbance travels through the medium as a wave, while particles only oscillate about their positions.
• This demonstrates a longitudinal wave, where particle motion is parallel to wave direction.
• Energy is transferred through the medium without actual movement of particles.

Viva Voce Questions

Question 1.
What is a disturbance in this activity?

Question 2.
What are compressions and rarefactions?

Question 3.
Does the marked point move along the slinky? Why?

Question 4.
What type of wave is formed in the slinky?

Question 5.
What is the difference between particle motion and wave motion?

Activity 5: Let us experiment (Musical Notes and Frequency)

Objective:

• To understand the relationship between frequency and musical notes.
• To observe how different notes have different frequencies.
• To identify patterns in the ratio of frequencies of musical notes.

Materials Required:

• Mobile phone with a frequency analyzer app (e.g., Phyphox)
• Optional: Another phone or musical instrument for generating notes

Procedure:

1. Open a mobile app that can measure sound frequency (Audio Spectrum).
2. Produce musical notes Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa one by one (by singing or using an app).
3. Observe and record the frequency shown for each note.
4. Compare the frequencies of all notes by taking the ratio with respect to ‘Sa’.
5. If possible, compare frequencies produced by voice and mobile app for the same notes.

Observation:

• Each musical note has a different frequency.
• The frequency is lowest for ‘Sa’ and gradually increases for higher notes.
• The ratios between frequencies of notes follow a definite pattern.
• Slight variation may occur between voice and digital sound.

Conclusion:

• Musical notes differ due to their frequency values.
• Higher frequency corresponds to a higher pitch.
• There exists a fixed relationship (ratio) between the frequencies of musical notes.

Viva Voce Questions

Question 1.
What is frequency?

Question 2.
Which musical note has the lowest frequency?

Question 3.
How does frequency affect pitch?

Question 4.
Why do different notes sound different?

Question 5.
What pattern do you observe in the frequencies of musical notes?

Activity 6: Let us experiment (Hearing Range of Humans)

Objective:

• To understand the range of human hearing.
• To observe how frequency affects sound perception.
• To identify audible, infrasonic, and ultrasonic sounds.

Materials Required:

• Mobile phone with a sound frequency generator app
• Earphones (optional)

Procedure:

1. Open a mobile app that can generate sound frequencies.
2. Set the frequency to 100 Hz, play the sound, and listen carefully.
3. Gradually increase the frequency in steps of 100 Hz up to 1000 Hz and observe the change in sound.
4. Now set the frequency to 50 Hz and slowly reduce it to around 20 Hz or until the sound is no longer heard.
5. Note the range of frequencies that you can hear clearly.

Observation Table:

• Sound is clearly heard within a certain frequency range.
• As frequency increases, the pitch of sound increases.
• At very low frequencies (below ~20 Hz), sound becomes inaudible.
• Different individuals may hear slightly different ranges.

Conclusion:

• Humans can hear sounds roughly in the range of 20 Hz to 20,000 Hz (20 kHz).
• Sounds below 20 Hz are called infrasonic, and above 20 kHz are ultrasonic.
• The ability to hear varies with age and individual sensitivity.

Viva Voce Questions

Question 1.
What is the audible range of human hearing?

Question 2.
What happens to pitch when frequency increases?

Question 3.
What are infrasonic waves?

Question 4.
What are ultrasonic waves?

Question 5.
Why do different people hear different frequency ranges?

Start by reviewing these Exploration Class 9 Science Solutions Chapter 8 Journey Inside the Atom Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 8 Question Answer

Class 9 Science Ch 8 Journey Inside the Atom Question Answer

Journey Inside the Atom Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
Choose the correct options and explain the reason for the correct and incorrect options in the context of Ernest Rutherford’s gold foil experiment.
(i) The experiment clearly showed the existence of neutrons in the nucleus.
(ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom.
(iii) The large deflection of a few alpha particles indicated that most of the mass of the atom and positive charge are packed into a tiny centre.
(iv) The way alpha particles were deflected showed that electrons move around the nucleus.
Solution:
The correct options are (ii) and (iii).

Reason for correct options:
During the experiment, a few heavy alpha particles bounced back sharply. This proved that the atom’s positive charge and mass are not spread out evenly (which disproves Thomson’s plum pudding model). Instead, they are tightly packed into a tiny, dense centre called the nucleus.

Reason for incorrect options:
Option (i) is incorrect because neutrons were discovered much later by James Chadwick, not by Rutherford’s experiment. Option (iv) is incorrect because the deflection of alpha particles was caused by the heavy, positive nucleus, not by the very light electrons.

Question 2.
Which of the following statements are correct or incorrect according to the Bohr’s atomic model? Give a reason for each statement.
(i) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus.
(ii) Electrons can exist anywhere around the nucleus with no fixed energy.
(iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy.
(iv) Electrons can be found between energy levels as they move around the nucleus.
Solution:
(i) is incorrect.
According to Bohr, electrons revolve in special stationary orbits and do not lose energy while moving in them.

(ii) is incorrect.
Electrons cannot exist just anywhere. They must revolve only in specific, allowed circular paths called energy shells.

(iii) is correct.
This is the main principle of Bohr’s model. Electrons revolve in fixed energy shells (like K, L, M) and their energy remains constant as long as they stay in that shell.

(iv) is incorrect.
Electrons cannot stay in the empty spaces between the energy levels. They can jump from one level to another, but cannot be found in between.

Question 3.
The composition of the nuclei of three atomic species X, Y, and Z are given as follows

	X	Y	Z
Number of protons	18	17	17
Number of neutrons	19	18	20

Explain the relation between the following,
(i) Y and Z
(ii) Z and X.
Solution:
(Based on the data provided for these species: X has 18p,19n; Y has 17p, 18n; Z has 17p, 20n).

(i) Relation between Y and Z: Both Y and Z have exactly the same number of protons (17), meaning they belong to the same element (Chlorine). However, they have a different number of neutrons (18 and 20). Therefore, Y and Z are isotopes.

(ii) Relation between Z and X: If we calculate their mass numbers (protons + neutrons), X has a mass number of 37 (18 + 19), and Z also has a mass number of 37 (17 + 20). Since they have the same mass number but different atomic numbers (different protons), Z and X are isobars.

Question 4.
What conclusion did Rutherford draw about the position and characteristics of the atom’s positively charged part based on the few alpha particles that bounced back or were deflected at large angles in the gold foil experiment?
Solution:
According to Rutherford’s experiment, a few fast and heavy alpha particles bounced almost straight back, Rutherford concluded that the positive charge of an atom is not spread out like a soft cloud. Instead, he realised that all the positive charge and almost the entire heavy mass of the atom are tightly packed into a very tiny, hard, and dense region exactly at the centre of the atom. He named this central part the nucleus.

Question 5.
Explain and arrange the following statements in the correct chronological order to show how atomic models have evolved over time.
(i) Bohr’s model proposed that electrons move in fixed orbits around the nucleus, each with a definite energy.
(ii) Thomson’s model depicted the atom as a ‘plum pudding’ with electrons embedded in a sphere of positive charge.
(iii) Rutherford’s model proposed that atoms have a dense central nucleus.
(iv) Dalton’s model described atoms as indivisible particles.
Solution:
Correct order:
(iv) Dalton’s model
(ii) Thomson’s model
(iii) Rutherford’s model
(i) Bohr’s model

Question 6.
Electrons move around the nucleus in orbits. Why do they not fly away from the atom? Explain what keeps them attracted to the nucleus.
Solution:
Electrons do not fly away from the atom because they are held tightly by a strong electrostatic force of attraction. The nucleus contains protons, which are positively charged, while electrons are negatively charged. Since opposite charges naturally attract each other, this strong inward pull from the positive nucleus keeps the fast-moving negative electrons perfectly in their orbits.

Question 7.
Assertion (A): The discovery of subatomic particles helped in understanding the atomic structure.
Reason (R): The number of electrons is equal to the number of protons in an atom.
Choose the correct option.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
The correct option is (ii) Both A and R are true, but R is not the correct explanation of A. Discovering electrons, protons and neutrons helped scientists build better atomic models over time.

Also, it is true that a neutral atom has equal electrons and protons. However, the equal number of charges (Reason) is not the actual explanation for how discovering these particles helped to understand the atomic structure (Assertion).

Question 8.
Magnesium is essential for many biological processes, including muscle contraction. For an atom of magnesium with a mass number of 24 and atomic number 12, determine the number of (i) protons, (ii) neutrons, (iii) electrons, and also illustrate the arrangement of electrons in a magnesium atom.
Solution:
Mass number of magnesium = 24
Atomic number of magnesium = 12
(i) Number of Atomic protons = 12
(ii) Number of neutrons = Mass number – Number of protons
Number of neutrons = 24 – 12 = 12
(iii) Number of electrons = Number of protons = 12
Electronic configuration = 2, 8, 2

Question 9.
Find the following information for the elements shown in the figure given below.
(i) Name of the element
(ii) Symbol
(iii) Total number of electrons
(iv) Number of valence electrons
(v) Valency of the element
(vi) Number of protons
(vii) Atomic number

Solution:
(a) Element: Lithium
Total electrons : 3
Valency: 1
Atomic number : 3
Symbol : Li
Valence electrons: 1
Protons : 3

(b) Element: Nitrogen
Total electrons : 7
Valency: 3
Atomic number : 7
Symbol : N
Valence electrons : 5
Protons : 7

(c) Element: Aluminium
Total electrons : 13
Valency: 3
Atomic number : 13
Symbol : Al
Valence electrons : 5
Protons : 13

(d) Element: Fluorine
Total electrons : 9
Valency: 1
Atomic number : 9
Symbol : F
Valence electrons : 7
Protons : 9

Question 10.
Both Rutherford’s and Bohr’s models have electrons orbiting the nucleus. Why did Rutherford’s model fail to explain atomic stability, while Bohr’s model succeeded?
Solution:
Rutherford’s model failed because, any charged particle moving in a circular path should constantly lose energy in the form of radiation. If this happened, the electron would lose its energy, spiral inward, and crash into the nucleus, making the atom ustable. Bohr’s model succeeded because he introduced a new rule: electrons revolve in special “stationary orbits.” He stated that as long as an electron stays in its fixed orbit, it simply does not radiate or lose any energy, which perfectly explains why the atom is stable.

Question 11.
An atom ⁷⁰X has 31 electrons. How many neutrons are there in its nucleus?
Solution:
A normal atom is electrically neutral, so the number of protons exactly equals to the number of electrons. So, the atom has 31 protons.
Mass Number = Protons + Neutrons.
To find neutrons, we reverse the formula,
Neutrons = Mass Number – Protons.
Neutrons = 70 – 31 = 39.
Thus, there are 39 neutrons in its nucleus.

Question 12.
An atom has 79 protons and a mass number of 197. Calculate
(i) the number of neutrons, and
(ii) the number of electrons.
Solution:
(i) Number of neutrons :
Neutrons = Mass Number – Protons.
So, 197 – 79 = 118 neutrons.

(ii) Number of electrons: An atom is electrically neutral, so the number of electrons equals to the number of protons. Thus, it has 79 electrons.

Question 13.
Complete the table given below.

Solution:
To fill out the table, we use simple atomic rules:
Atomic Number = Protons = Electrons; and
Mass Number = Protons + Neutrons.

Atomic No.	Mass No.	Neutrons	Protons	Electrons	Name of Element
5	11	6	5	5	Boron
7	14	7	7	7	Nitrogen
12	24	12	12	12	Magnesium
15	31	16	15	15	Phosphorus
1	1	0	1	1	Hydrogen

Question 14.
A man was discussing the structure of atom with his classmates. During the discussion, he learnt that an element X has a mass number of 35 and contains 18 neutrons. Based on this information, answer the following questions.
(i) How many electrons and protons does element X have?
(ii) What is its atomic number?
(iii) Identify the element X.
(iv) Write its electronic configuration.
(v) How many valence electrons does it have?
(vi) What will be the mass number if two neutrons are added to its nucleus?
(vii) What will be the relation of X with the new atom?
Solution:
(i) Protons = Mass Number – Neutrons
= 35 -18 =17.
So, it has 17 protons and 17 electrons.

(ii) The atomic number is always equal to the number of protons, so it is 17.

(iii) The element with the atomic number 17 is chlorine (Cl).

(iv) With 17 electrons, shells are filled as K (2), L (8), M (7).
The configuration is 2, 8, 7.

(v) Looking at the outermost shell, it has 7 valence electrons.

(vi) If two neutrons are added, the new mass number becomes 35 + 2 = 37.

(vii) Since the number of protons remains the same (17) but the mass number has changed (37), the new atom will be an isotope of element X.

Question 15.
In an atom, there are 12 protons and 12 neutrons in the nucleus. Now, imagine that all the electrons are replaced with some hypothetical particles that have the same charge as electrons but are 500 times heavier. What effect will this replacement have on the atom’s?
(i) Atomic number
(ii) Atomic mass
(iii) Mass number
(iv) Overall charge
Solution:
(i) The atomic number will remain the same (12) because it depends only on the number of protons.
(ii) The atomic mass will increase because the new particles are much heavier than electrons.
(iii) The mass number will remain the same (24) as it depends only on the number of protons and neutrons.
(iv) The overall charge will remain neutral because the number of negatively charged particles is equal to the number of protons.

Class 9 Science Chapter 8 Journey Inside the Atom Question Answer (InText)

Pause and Ponder

Question 1.
Suppose you made up your own ‘atom’, as Thomson described, using clay for the positive charge and small beads for the electrons spread through it. What will happen if
(i) the positive charge on the clay is lesser than the total negative charge of the beads?
(ii) by mistake, the clay itself carries a bit of negative charge? Would your model still represent a neutral atom?
Solution:
(i) If the positive charge of the clay is less than the total negative charge of the beads, the extra negative charge will increase on atom. As a result, the entire atom will become negatively charged instead of being neutral.

(ii) If the clay itself carries a bit of negative charge along with the positive charge, it will add up with the negative charge of the beads. This means the total negative charge will become greater than the positive charge. Therefore, the model will no longer represent a neutral atom.

Question 2.
Could an orange or a lemon, which also contain seeds inside soft pulp, be a good comparison? In what ways does it match Thomson’s idea and where does it fall short?
Solution:
Yes, an orange or a lemon can be a fairly good comparison to Thomson’s model. It matches the idea correctly because the soft pulp is spread everywhere like the positive charge, and the seeds are spread inside just like the negatively charged electrons.

However, it falls short because the seeds in a lemon are mostly gathered near the center, whereas Thomson suggested that electrons are spread uniformly everywhere. Also, this fruit comparison does not show the true empty space that actually exists inside a real atom.

Question 3.
Why did Thomson conclude that electrons are present in all atoms?
Solution:
Thomson concluded this because, during his famous cathode ray tube experiment, he noticed that the nature of the cathode rays remained exactly the same even when he changed the gas inside the tube or used different materials for the electrodes. Since these tiny negatively charged particles (electrons) appeared exactly identical no matter what material was used, he then realised that they must be a basic, universal part of all atoms.

Question 4.
What do you think would happen if alpha particles were replaced with negatively charged particles in Rutherford’s gold foil experiment?
Solution:
If negatively charged particles, like electrons, were used instead of the positively charged alpha particles, they would behave very differently. Because opposite charges attract, these negative particles would be strongly attracted to the positive nucleus instead of being pushed away or deflected. Furthermore, since electrons are incredibly light compared to heavy alpha particles, they would easily lose their path and get captured by the atoms, completely changing the results of the experiment.

Question 5.
Rutherford found that a few a-particles bounced back sharply. How does this single surprising result completely rule out Thomson’s ‘plum pudding model’ of the atom?
Solution:
According to Thomson’s plum pudding model, the positive charge and mass were spread out evenly and very thinly like a soft cloud. If this were true, the heavy and fast alpha particles should have easily passed straight through the gold foil.

The fact that a few alpha particles hit something hard and bounced straight back proved that all the positive charge and heavy mass must be packed tightly into a tiny, dense center. This dense center (the nucleus) is the reason which hit a-rays back so hard, which completely disproved Thomson’s idea of a plum pudding model.

Question 6.
If you could ask Rutherford one question about his work, what would it be?
Solution:
If I could ask Rutherford one question, I would ask: “When you saw those alpha particles bounce straight back at you like a bullet hitting tissue paper, what was the very first thought that came to your mind before you discovered it was a dense nucleus?”

Question 7.
Assertion (A) Rutherford concluded that most of the mass of an atom is concentrated in a small region at the centre called the nucleus.
Reason (R) According to Thomson’s model, electrons are embedded in a uniformly distri¬buted positive charge sphere.
Choose the correct option.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Solution:
The correct option is (ii) Both A and R are true, but R is not the correct explanation of A.

Rutherford discovered the nucleus, and Thomson proposed the uniform positive sphere. However, Rutherford discovered the heavy central mass because heavy alpha particles bounced back during his gold foil experiment, not because of what Thomson’s earlier model had stated.

Question 8.
Imagine you are a scientist who has discovered a new element. Name this element after yourself and justify that the symbol you have chosen follows the IUPAC rules.
Solution:
Let us imagine my name is “Ayan,” and I proudly name my newly discovered element “Ayanium.” According to the IUPAC rules, I will choose the symbol “Ay”. To justify this, the first letter ‘A’ is written in capital (uppercase), and the second letter ‘y’ is written in small (lowercase), following the same uppercase-lowercase rule, which makes it scientifically correct.

Question 9.
What problems could arise if every scientist used different symbols for the same element?
Solution:
If different scientists used different symbols for the same element, it would lead to confusion and lack of uniformity, making scientific communication and understanding difficult.

Question 10.
An atom with an atomic number of 26 has 56 nucleons. Find out its number of electrons, protons and neutrons.
Solution:

• The atomic number of an element is the same as the number of protons in that element. So, it has 26 protons.
• Since a normal atom is neutral, the number of electrons is equal to the number of protons. Thus, it has 26 electrons.
• The term “nucleons” refers to the total number of protons and neutrons together (Mass Number = 56).
• To find the neutrons, we subtract the protons from the mass number: 56 – 26 = 30. Therefore, the atom has 30 neutrons.

Question 11.
The nucleus of an atom contains 20 protons. If its mass number is 41, find the number of neutrons in it.
Solution:

• The mass number of an atom is the sum of its protons and neutrons. It is given that the mass number is 41 and the number of protons is 20.
• To find the number of neutrons, we subtract the protons from the mass number: 41 – 20 = 21. Therefore, the atom contains 21 neutrons.

Question 12.
An atom has 18 neutrons and an atomic number of 17. What is its mass number?
Solution:

• The atomic number tells exactly how many protons any atom has. Here, the atomic number is 17, which means there are 17 protons.
• The mass number is calculated by adding the number of protons and the number of neutrons together.
• So, Mass Number = 17 + 18 = 35.
• Therefore, the mass number of this atom is 35.

Question 13.
An atom ²⁴A has 11 electrons. Find the number of neutrons in it.
Solution:

• The general symbol of an element is %A. For any neutral atom, no. of protons = no. of electrons.
• So, the number of protons is 11, and the mass number is 24.
• To find the number of neutrons, we subtract the number of protons from the mass number: 24 – 11 = 13
• Therefore, there are 13 neutrons in this atom.

Question 14.
Identify the number of electrons in the outermost shell of the following elements
(i) \({ }_{6}^{12} C\)
(ii) \({ }_{9}^{19} F\)
(iii) \({ }_{14}^{28} Si\)
Solution:
(i) Carbon (\({ }_{6}^{12} C\)) :
The atomic number (bottom number) is 6, meaning it has 6 electrons. Its electronic configuration is 2, 4. Therefore, it has 4 electrons in its outermost shell.

(ii) Fluorine (\({ }_{9}^{19} F\)) :
The atomic number is 9, meaning it has 9 electrons. Its electronic configuration is 2, 7. Therefore, it has 7 electrons in its outermost shell.

(iii) Silicon (\({ }_{14}^{28} Si\)) :
The atomic number is 14, meaning it has 14 electrons. Its electronic configuration is 2,8, 4. Therefore, it has 4 electrons in its outermost shell.

Question 15.
Write the electronic configuration of the elements having atomic numbers 12, 16 and 18.
Solution:
To write the electronic configurations, we fill the shells in the order of K (2), L (8) and M (18).

• For atomic number 12 (Magnesium), the electrons are distributed as 2, 8, 2.
• For atomic number 16 (Sulphur), the electrons are distributed as 2, 8, 6.
• For atomic number 18 (Argon), the electrons are distributed as 2, 8, 8.

Question 16.
Solve this riddle I am an atom with a mass number of 23 and 11 protons. I am a soft metal and react vigorously with water. Who am I and how many neutrons do I have? You can also create one such riddle.
Solution:
Since you have exactly 11 protons, your atomic number is 11. The element with atomic number 11 is Sodium (Na), which is indeed a soft metal that reacts strongly with water. To find neutrons, I will subtract the protons from the mass number: 23 – 11 = 12. You have 12 neutrons.

My Riddle :
I am an atom with 8 protons and a mass number of 16. I am an invisible gas that all living beings breathe in to stay alive. Who am i and what is my electronic configuration?
(Answer: Oxygen and its electronic configuration is 2, 6).

Question 17.
Two different atoms have 11 protons each, but one has 12 neutrons, and the other has 13 neutrons. How do their atomic numbers and mass numbers compare? Are they the same element or different elements?
Solution:
Since both atoms have 11 protons, their atomic number is the same (11).
Mass number = Number of protons + Number of neutrons
First atom: 11 + 12 = 23
Second atom: 11 + 13 = 24
Therefore, both atoms have the same atomic number but different mass numbers (23 and 24).
Hence, they are isotopes of the same element, not different elements.

Question 18.
If a bromine atom is available in the form of, say, two isotopes, \({ }_{35}^{79} Br\) (49.7 %) and \({ }_{35}^{81} Br\) (50.3 %), calculate the average atomic mass of the bromine atom.
Solution:
Isotopes of bromine:
₃₅⁷⁹ Br = 49.7%
₃₅⁸¹ Br = 50.3%
Average atomic mass is calculated using

weighted average:
Average atomic mass = \(\frac{79 \times 49.7}{100}+\frac{81 \times 50.3}{100}\)
Now calculating:
79 × 49.7 = 3926.3
81 × 50.3 = 4074.3
Total = 3926.3 + 4074.3 = 8000.6
Average atomic mass 8.000.6 ÷ 100 = 80.006 u

Start by reviewing these Exploration Class 9 Science Solutions Chapter 7 Work Energy and Simple Machines Question Answer NCERT Solutions to strengthen your knowledge.

Class 9 Science Exploration Chapter 7 Question Answer

Class 9 Science Ch 7 Work Energy and Simple Machines Question Answer

Work Energy and Simple Machines Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine

Question 1.
State whether True or False.

(i) Work is said to be done when a force is applied, even if the object does not move.
Solution:
False, work requires displacement. If the object does not move, work done is zero.

(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
Solution:
True, both the applied force and the displacement are in the upward direction.

(iii) The SI unit for both work and energy is joule (J).
Solution:
True.

(iv) A motionless stretched rubber band has kinetic energy.
Solution:
False, a motionless stretched rubber band possesses elastic potential energy, not kinetic energy.

(v) Energy can change from one form to another.
Sol
True.

Question 2.
Fill in the blanks.

(i) Work done = _____________ × _____________ (in the direction of force).
Solution:
Work done = force × displacement (in the direction of force).

(ii) 1 joule of work is done when a force of _____________ newton displaces an object by 1 metre in the direction of the force.
Solution:
1 joule of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.

(iii) The expression for kinetic energy of a body of mass m and velocity v is _____________ .
Solution:
The expression for kinetic energy of a body of mass m and velocity v is 1/2 mv².

(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is _____________.
Solution:
The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.

(v) Power is defined as the _____________ at which work is done.
Solution:
Power is defined as the rate at which work is done.

Question 3.
When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
(i) The force acting on the ball is zero.
(ii) The acceleration of the ball is zero.
(iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.
Solution:
Correct statements are (iii) Its kinetic energy is zero and (iv) Its potential energy is maximum.

At the highest point, the velocity momentarily becomes zero, so kinetic energy is zero. Since it is at its maximum height, potential energy is at its peak. Gravity (force and acceleration) continues to act downwards.

Question 4.
For each of the following situations, identify the energy transformation that takes place
(i) a truck moving uphill,
(ii) unwinding of a watch spring,
(iii) photosynthesis in green leaves,
(iv) water flowing from a dam,
(v) burning of a matchstick,
(vi) explosion of a fire cracker,
(vii) speaking into a microphone,
(viii) glowing electric bulb, and
(ix) solar panel.
Solution:
(i) Truck moving uphill: Chemical energy → Kinetic + Potential + Heat energy.

(ii) Unwinding of a watch spring: Elastic potential energy → Kinetic energy.

(iii) Photosynthesis in green leaves: Light energy → Chemical energy.

(iv) Water flowing from a dam: Potential energy → Kinetic energy.

(v) Burning of a matchstick: Chemical energy → Heat energy + Light energy.

(vi) Explosion of a fire cracker: Chemical energy → Sound + Light + Heat + Kinetic energy.

(vii) Speaking into a microphone: Sound energy → Electrical energy.

(viii) Glowing electric bulb: Electrical energy → Light energy + Heat energy.

(ix) Solar panel: Light energy → Electrical energy.

Question 5.
A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g =10 ms^-2 and student’s mass is m = 50 kg.
(i) Find the gain in the potential energy if the student is lifted straight up to the top.
(ii) Find the gain in the potential energy when the student climbs the stairs to the same top.
(iii) What do you conclude about the dependence of the potential energy on the path taken?
Solution:
(i) Gain in potential energy (lifted straight up),
∆ U = mgh
= 50 kg × 10 ms^-2 × 72.5 m
= 36250 J

(ii) Gain in potential energy (climbing stairs),
∆ U = mgh
= 50kg × 10 ms^-2 × 72.5 m
= 36250 J

(iii) The change in potential energy depends only on the initial and final vertical positions (height), not on the path taken.

Question 6.
A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Solution:
Let the height of one floor be h and the time taken for the 10th floor be t.
Energy:
The 20th floor is twice as high as the 10th floor (20 h vs 10 h).
E₁₀ = mg (10h)
E₂₀ = mg(20 h) = 2 × E₁₀
The energy required for the 20th floor is double.

Power:
Power is the rate of doing work.
P₁₀ = \(\frac{m g(10 \mathrm{~h})}{t}\)
P₂₀= \(\frac{m g(20 \mathrm{~h})}{2 t}\)
= \(\frac{m g(10 \mathrm{~h})}{t}\)
= P₁₀
The power required remains the same.

Question 7.
Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
Solution:
The energy (work) required is determined by the weight of the flag (mg) and the height (h) of the pole (W = mgh).
Raising it slowly or quickly does not change the total work done, as work is independent of time.
However, Power = Work / Time. If speed is doubled, the time is halved.
Therefore, the power requirement is doubled.

The factors determining the energy required are the mass of the flag, the height of the flagpole and the acceleration due to gravity.
Raising the flag slowly or quickly does not change the amount of work done, as work depends only on force and vertical displacement (W = mgh).
If the speed is doubled, the time taken is halved (t → \(\frac{t}{2}\)).
P_new = \(\frac{W}{t/2}\)
= 2 (\(\frac{W}{t}\))
= 2 P_old
Therefore, the power requirement is doubled.

Question 8.
A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Solution:
Day 1: Mass = 100 kg (scooter) + 60 kg (man) = 160 kg.
Day 2:160 kg + 40 kg = 200 kg.
Both days reach same speed v from rest in same time:
KE at Day 1 = \(\frac{1}{2}\) × 160 × v² = 80v².
KE at Day 2 = \(\frac{1}{2}\) × 200 × v² = 100 v².
Fuel ratio = 80 v² : 100 v² = 4: 5

Question 9.
On a see-saw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The see-saw, however, is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Solution:
To balance a see-saw, the product of the force (weight) and its distance from the fulcrum must be equal on both sides. This is known as the principle of the lever.
Let the weight of the child be W.
Then the weight of the adult is 2 W.
If the child sits at a distance d from the fulcrum, the adult must sit at a distance x such that
W × d = 2 W × x
x = \(\frac{d}{2}\)
Therefore, the adult must sit at half the distance from the fulcrum compared to the child.

Question 10.
A ball of mass 2 kg is thrown up with a velocity of 20 ms^-1.
(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.
(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 ms^-2).
Solution:
(i) Upward motion: negative (force of gravity is opposite to the direction of displacement).
Downward motion: positive (force of gravity is in the same direction as displacement).

(ii) Work done by air resistance equals the change in total mechanical energy.
K_i = ½ mv²
= ½ × 2kgx (20 ms^-1)²
= 400 J

U_i = mgh
= 2 kg × 10 ms^-2 × 19.4 m
= 388 J

W_air = U_f – K_i
= 388 J – 400 J = – 12 J

Question 11.
A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the figure, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed
(i) at 0 m, and
(ii) at 4 m.
Does the block have negative acceleration in any portion of its motion?

Solution:
Total work done is the area under the force – displacement graph.
W = ½ × (Sum of parallel sides) × Height
= ½ × (4m + 2m) × 50 N = 150 J

(i) Speed at 0 m,
½ m\(\ v_0^2\) = 180 J
½ × 10 × \(\ v_0^2\) =180
⇒ v₀ = \(\sqrt{36}\) = 6 ms^-1

(ii) Speed at 4 m (Work-Energy Theorem),
W = K_f – K_i
150 = K_f – 180
⇒ K_f = 330 J
½ × 10 × \(\ v_f^2\) = 330
⇒ v_f = \(\sqrt{66}\)≈ 8.12 ms^-1
The block does not have negative acceleration because the applied force is always positive or zero.

Question 12.
The gravitational attraction on the surface of the Moon (lunar surface) is about 1 /6 th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Solution:
On Earth: Ball thrown up to height h = 8 m with velocity u.
Using v² = u² – 2g_earth × h,
u² = 2 × 10 × 8 = 160 m² S^-2
On Moon: g_moon = \(\frac{g_{\text {earth }}}{6}=\frac{10}{6} \)= ms^-2 ≈ 1.67 ms^-2
Height on Moon: h_moon = \(\frac{u^2}{\left(2 \times g_{\text {moon }}\right)}=\frac{160}{\left(2 \times \frac{10}{6}\right)} \)
160 × \(\frac{6}{20}\) =48m
The ball will travel up to 48 m on the Moon with the same initial upward velocity.

Question 13.
A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of the motion of the car starting from the instant the driver spots the traffic ahead is shown in the figure.
(i) Describe how the car moves between positions A and B.
(ii) Calculate the kinetic energy of the car at A
(iii) State the work done by the brakes in bringing the car to a halt between B and C.
(iv) What does the kinetic energy of the car transform into?

Solution:
(i) Between A and B, the car moves at a constant speed of 35 m/s^-1 (driver’s reaction time).

(ii) Kinetic energy at A,
K_A = ½ mv²
= ½ × 1000 kg (35 ms^-1)²
= – 612500 J

(iii) Work done by brakes = Change in Kinetic Energy W
= K_C – K_B
= 0 – 612500 J
= – 612500 J

(iv) The kinetic energy transforms primarily into heat energy (friction) and sound energy.

Question 14.
The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in the figure. At O, the velocity of the ball is 0 m s”1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

Solution:
At x = 0 (Point O),
v = 0 ms^-1,
So, kinetic energy K = 0.
Total mechanical energy E = U + K
= 30 J + 0 J = 30 J.
Energy is conserved.

At P, U_P = 20 J
K_P = E – U_P
= 30 J – 20 J = 10 J
½ m\(\ v_P^2\) = 10 J
⇒ ½ (0.5 kg) \(\ v_P^2\) = 10
v_P = \(\sqrt{40}\) ≈ 6.32 ms^-1

At Q, U_Q = 30 J
K_Q = 30 J – 30 J = 0 J
⇒ v_Q = 0 ms^-1

At R, U_R =40 J.
The total energy is only 30 J, so the ball does not have enough energy to reach this height. The ball cannot reach point R.

Question 15.
A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.
(i) Calculate the velocity of the coconut just before it hits the sand.
(ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 ms^-2.
Solution:
Mass = 1.5 kg, height = 10 m, g = 10 ms^-2.
(i) Velocity just before hitting the sand:
Using conservation of energy: \(\frac{1}{2}\) mv² = mgh, v = \(\sqrt{2 \mathrm{gh})}=\sqrt{(2 \times 10 \times 10)} \) = \(\sqrt{200} \) ≈ 1.41 ms^-1

(ii) Depth of depression: All energy used to create depression.
Resistive force of sand = 3000 N.
Work done by sand = KE of coconut just before impact = mgh
= 1.5 × 1o × 10 = 150 J
⇒ 3000 N × Depth = 150J
Depth = 150/3000 = 0.05 m = 5 cm

Class 9 Science Chapter 7 Work Energy and Simple Machines Question Answer (InText)

Pause and Ponder

Question 1.
In the previous chapter, a weightlifter is shown holding a barbell steady in her hands. Is she doing any work on the barbell while holding it steady?

Solution:
No, in physics, work is done only when a force causes a displacement. While the weightlifter is applying a continuous upward force to counteract gravity, the barbell is held steady, meaning the displacement is zero (s = 0). Therefore, the work done on the barbell (W = F.s) is zero.

Question 2.
Is the work done by friction on the stack of coins that travels on a rough surface as shown in the figure positive, negative or zero?

Solution:
Negative. Friction acts opposite to the direction of motion (displacement). Since force and displacement are in opposite directions, work done by friction is negative.

Question 3.
When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Solution:
The muscular energy is primarily transformed into the kinetic energy of the moving bicycle and rider. Additionally, a portion of it is dissipated as heat energy (due to internal friction in the bicycle’s mechanical parts and tyre, as well as body heat generated by the rider).

Question 4.
Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Solution:
Moving in the horizontal direction does not change the object’s height relative to the Earth’s surface, so its gravitational potential energy remains constant.
If it is gradually raised in the vertical direction, its height increases, so its gravitational potential energy increases (U = mgh).

Question 5.
For the situation depicted in the figure, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.

Solution:
By the law of conservation of mechanical energy (ignoring air resistance), the total mechanical energy of an isolated system remains conserved. At the highest point, the energy is entirely potential (E = mgh).

As it falls, this converts to kinetic energy. Just before it hits the ground, height is zero (U = 0), but the kinetic energy equals the original potential energy. Thus, the total mechanical energy is still mgh, completely in the form of kinetic energy.

Question 6.
You may have seen an exhibit like that in figure in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Solution:
As the ball rolls from A to B, potential energy decreases and kinetic energy increases. From B to C, kinetic energy decreases and potential energy increases. Subsequent peaks (C, D, E) must be lower than A because total mechanical energy is not perfectly conserved in reality, a fraction of the energy is continually transformed into heat and sound due to friction and air resistance.

Question 7.
Why are roads on hills built to wind around in gentle slopes rather than going straight up?

Solution:
A winding road acts as an inclined plane with a very large length (L). Since MA = \(\frac{L}{h}\), increasing the length increases the mechanical advantage. This reduces the magnitude of the effort force the vehicle’s engine must provide, preventing it from stalling.

Question 8.
To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder. Explain why.

Solution:
A vertical ladder requires lifting your entire body weight straight up (MA = 1). An inclined ladder acts as an inclined plane, which increases the distance required to travel for the same work, thus reducing the force required. The mechanical advantage is greater than 1, so the muscular effort required is reduced.

Question 9.
Why is it easier to open the lid of a can using a spoon as shown in figure?
Solution:
The spoon acts as a Class I lever. The rim of the can acts as the fulcrum. Applying effort at the long end of the spoon creates a larger effort arm. Since MA = effort arm/load arm> 1, a small force at the end of the spoon exerts a much larger force to pop the lid off.

Question 10.
Why do you push an object closer to the fulcrum of a pair of scissors when you want to cut a hard object?
Solution:
Moving the object closer to the fulcrum decreases the length of the load arm. Since MA = Effort arm/Load arm, decreasing the denominator drastically increases the mechanical advantage, applying a much stronger cutting force on the object.

Question 11.
Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.

Solution:
All real machines slow down and stop because of energy losses. When a machine works, it uses its energy to do work. At the same time, some energy is always lost due to friction, air resistance, etc. This lost energy is mainly converted into heat (and sometimes sound), which cannot be fully reused. So, the machine’s total usable energy keeps decreasing. Eventually, no energy is left to do work, and the machine stops. That’s why perpetual motion machines are impossible-they would require no energy loss, which never happens in real life.

Class 9 Science Chapter 7 Question Answer (Activities)

Activity 1: Investigation of Potential Energy

Objective:

• To understand the concept of potential energy.
• To observe how height affects energy.
• To relate energy to the effect produced (depression in sand).

Materials Required:

• A heavy ball
• A large container filled with loose sand
• Measuring scale (optional)

Procedure:

1. Take a container filled with loose sand and place it on a flat surface.
2. Hold the ball at a height of about 1 metre above the sand.
3. Drop the ball and observe the depression formed in the sand.
4. Now raise the ball to a height of about 2 metres.
5. Drop it at a different spot so that depressions do not overlap.
6. Repeat the experiment for different heights.
7. Compare the depths of the depressions formed.

Observation:

• A depression is formed each time the ball falls on the sand.
• The depth of the depression increases with increase in height.
• The depression is:
  • Shallow at lower height
  • Deep at greater height

Conclusion:

• The ball at a greater height possesses more potential energy.
• When released, this energy converts into kinetic energy, producing a deeper depression.
• Therefore, potential energy increases with height.

Viva Voce Questions

Question 1.
What is potential energy?

Question 2.
On what factors does potential energy depend?

Question 3.
Why does the ball create a deeper depression from greater height?

Question 4.
What type of energy conversion takes place in this activity?

Question 5.
What is the SI unit of potential energy?

Activity 2: Demonstration of Conservation of Mechanical Energy (Simple Pendulum)

Objective:

• To understand the concept of kinetic energy (KE) and potential energy (PE).
• To demonstrate the conservation of mechanical energy.
• To observe energy transformation during oscillatory motion.

Materials Required:

• A simple pendulum (thread and bob)
• Stand or support
• White sheet of paper
• Pencil/marker

Procedure:

1. Set up a simple pendulum using a thread and a small bob.
2. Paste a white sheet behind the pendulum and draw a horizontal reference line at the level of the bob when it is at rest.
3. Pull the bob to one side (point P) up to the horizontal line.
4. Release the bob and allow it to oscillate freely.
5. Observe the motion at: Extreme points (P and R), Mean position (Q)
6. Note the height reached on the opposite side and compare it with the initial height.

Observation:
At extreme positions (P and R):

• The bob momentarily comes to rest.
• Kinetic Energy = 0, Potential Energy = Maximum.

At the mean position (Q):

• The speed of the bob is maximum.
• Kinetic Energy = Maximum, Potential Energy = Minimum (~ 0).
• The bob reaches almost the same height on both sides.

Conclusion:

• Energy continuously transforms between potential energy and kinetic energy.
• The total mechanical energy (KE + PE) remains constant during motion.
• This verifies the law of conservation of mechanical energy.
• In real conditions, slight energy loss occurs due to air resistance and friction.

Viva Voce Questions

Question 1.
What forms of energy are involved in a pendulum?

Question 2.
At which point is kinetic energy maximum?

Question 3.
Why is potential energy maximum at extreme positions?

Question 4.
Why does the pendulum not reach the same height in real life?

Question 5.
State the law of conservation of mechanical energy.

Activity 3: Let us experiment (Inclined Plane and Force)

Objective:

• To understand how an inclined plane reduces the force required to lift an object.
• To study the relation between force and distance.
• To observe the effect of the angle of inclination on effort.

Materials Required:

• Smooth plank (cardboard/wooden board ~1.5 m)
• Toy cart
• Spring balance
• Stack of books or stool

Procedure:

1. Arrange a raised platform using books or a stool (about 0.5 m height).
2. Attach the spring balance to the toy cart.
   Step A (Vertical lifting)
3. Lift the cart vertically from the floor to the top of the stack slowly.
4. Note the reading on the spring balance.
   Step B (Inclined plane)
5. Place the plank as an inclined plane against the stack.
6. Pull the cart up the plank steadily using the spring balance.
7. Record the reading.
   Step C (Changing angle)
8. Reduce the angle of the plank (make it less steep).
9. Pull the cart again and note the new reading.

Observation:

• Maximum force is required when the cart is lifted vertically.
• Less force is required when pulling the cart on an inclined plane.
• As the plank becomes less steep, the required force decreases further.
• The distance travelled increases when the slope is less steep.

Conclusion:

• An inclined plane helps in reducing the effort (force) needed to lift objects.
• The reduction in force is achieved by increasing the distance over which the force acts.
• Thus, less force but more distance is required to reach the same height.

Viva Voce Questions

Question 1.
What is an inclined plane?

Question 2.
Why is less force required on a less steep incline?

Question 3.
What happens to the distance when the slope is reduced?

Question 4.
Compare the force required for vertical lifting and an inclined plane.

Question 5.
Give examples of inclined planes in daily life.